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Specialized Subject - STEM 2020 GENERAL PHYSICS 1 SCIENCE TECHNOLOGY ENGINEERING AND MATHEMATICS Specialized Subject 1/1/2020 0 Specialized Subject - STEM MODULE1Q1 MEASUREMENTS Lesson 1: Units of Measurements Overview: This module was designed and written with you in mind. It is here to help you master the Units and Measurements. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Solve measurement problems involving conversion of units, expression of measurements in scientific notation Lesson Learning Objectives: The learners are expected to be able to… ✓ ✓ ✓ ✓ ✓ ✓ define physical quantity; differentiate fundamental and derive quantity; differentiate metric and British system of measurement; convert units of measurement; express number in scientific notation; and solve measurement problems involving conversion of units and expression in scientific notation PRE-TEST Direction: Read the questions properly, choose the correct answer and write it on the blanks provided before each item. ____1. How much wood do you need to a form a triangular garden frame if one side of the frame has a length of 11 ft, and the other two sides are 2 feet longer than the first side? a. 33 ft c. 36 ft b. 35 ft d. 37 ft ____2. How many inches is 9’10”? 1 Specialized Subject - STEM a. 116” c. 129” b. 118” d. 228” ____3. How many yards is 9 mi? a. 12 672 yards c. 15 840 yards b. 14 500 yards d. 16 040 yards ____4. Jessie measures her bathroom tiles to be 10 in by 8 in. What are the length and width in cm? a. 20.32 cm by 19.6 cm c. 25.4 cm to 24.5 cm b. 25.4 cm by 20.32 cm d. 35.4 cm by 12.32 cm _____5. Convert 5 761 millimeters to meters. a. 5.761 c. 576 100 b. 57.61 d. 5 761 000 PRIMING Physical Quantities All physical quantities in the International System of Units (SI) are expressed in terms of combinations of seven fundamental physical units, which are units for: length, mass, time, electric current, temperature, amount of a substance, and luminous intensity. SI Units: Fundamental and Derived Units There are two major systems of units used in the world: SI units (acronym for the French Le Système International d’Unités, also known as the metric system), and English units (also known as the imperial system). English units were historically used in nations once ruled by the British Empire. Today, the United States is the only country that still uses English units extensively. Virtually every other country in the world now uses the metric system, which is the standard system agreed upon by scientists and mathematicians. Some physical quantities are more fundamental than others. In physics, there are seven fundamental physical quantities that are measured in base or physical fundamental units: length, mass, time, electric current temperature, amount of substance, and luminous intensity. Units for other physical quantities (such as force, speed, and electric charge) described by mathematically combining these seven 2 Specialized Subject - STEM base units. In this course, we will mainly use five of these: length, mass, time, electric current and temperature. The units in which they are measured are the meter, kilogram, second, ampere, kelvin, mole, and candela. All other units are made by mathematically combining the fundamental units. These are called derived units. Table 1. SI Base Units Quantity Name Symbol Length Meter m Mass Kilogram kg Time Second s Electric current Ampere A Temperature Kelvin K Amount of substance Mole mol Luminous intensity Candela cd Metric Prefixes Physical objects or phenomena may vary widely. For example, the size of objects varies from something very small (like an atom) to something very large (like a star). Yet the standard metric unit of length is the meter. So, the metric system includes many prefixes that can be attached to a unit. Each prefix is based on factors of 10 (10, 100, 1,000, etc., as well as 0.1, 0.01, 0.001, etc.). Table 2 Metric Prefixes and symbols used to denote the different various factors of 10 in the metric system Prefix Symbol Value Exa E 1018 Peta P 1015 Example Name Example Symbol Example Value Example Description Distance light travels in a century Exameter Em 1018 Petasecond Ps 1015 s m 30 million years 3 Specialized Subject - STEM Example Symbol Example Value Example Description Terawatt TW 1012 W Powerful laser output 109 Gigahertz GHz 109 Hz A microwave frequency M 106 Megacurie MCi 106 Ci High radioactivity Kilo K 103 Kilometer Km 103 m About 6/10 mile hector H 102 Hectoliter hL 102 L 26 gallons Deka Da 101 Dekagram Dag 101 g Teaspoon of butter ____ ____ 100 (=1) Deci D 10–1 Deciliter dL 10–1 L Less than half a soda Centi C 10–2 Centimeter Cm 10–2 m Fingertip thickness Mili M 10–3 Millimeter Mm 10–3 m Flea at its shoulder Micro µ 10–6 Micrometer µm 10–6 m Detail in microscope Nano N 10–9 Nanogram Ng 10–9 g Small speck of dust Prefix Symbol Value Tera T 1012 Giga G Mega Example Name 4 Specialized Subject - STEM Prefix Symbol Value Pico P 10–12 Femto F 10–15 A 10–18 Atto Example Name Example Symbol Example Value Example Description Small capacitor in radio Picofarad pF 10–12 Femtometer Fm 10–15 m Size of a proton As 10–18 Time light takes to cross an atom Attosecond F s The metric system is convenient because conversions between metric units can be done simply by moving the decimal place of a number. This is because the metric prefixes are sequential powers of 10. There are 100 centimeters in a meter, 1000 meters in a kilometer, and so on. In nonmetric systems, such as U.S. customary units, the relationships are less simple—there are 12 inches in a foot, 5,280 feet in a mile, 4 quarts in a gallon, and so on. Another advantage of the metric system is that the same unit can be used over extremely large ranges of values simply by switching to the most-appropriate metric prefix. For example, distances in meters are suitable for building construction, but kilometers are used to describe road construction. Therefore, with the metric system, there is no need to invent new units when measuring very small or very large objects—you just have to move the decimal point (and use the appropriate prefix). PROCESSING Unit Conversion and Dimensional Analysis A conversion factor relating meters to kilometers. A conversion factor is a ratio expressing how many of one unit are equal to another unit. A conversion factor is simply a fraction which equals 1. You can multiply any number by 1 and get the same value. When you multiply a number by a conversion factor, you are simply multiplying it by one. For example, the following are conversion factors: 1 foot/12 inches = 1 to convert inches to feet, 1 meter/100 centimeters = 1 to convert centimeters to meters, 1 minute/60 seconds = 1 to convert seconds to minutes 5 Specialized Subject - STEM In this case, we know that there are 1,000 meters in 1 kilometer. Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor (1 km/1,000m) = 1, so we are simply multiplying 80m by 1: Using Scientific Notation with Physical Measurements Scientific notation is a way of writing numbers that are too large or small to be conveniently written as a decimal. For example, consider the number 840,000,000,000,000. It’s a rather large number to write out. The scientific notation for this number is 8.40 × 1014. Scientific notation follows this general format x × 10y In this format x is the value of the measurement with all placeholder zeros removed. In the example above, x is 8.4. The x is multiplied by a factor, 10y, which indicates the number of placeholder zeros in the measurement. Placeholder zeros are those at the end of a number that is 10 or greater, and at the beginning of a decimal number that is less than 1. In the example above, the factor is 10 14. This tells you that you should move the decimal point 14 positions to the right, filling in placeholder zeros as you go. In this case, moving the decimal point 14 places creates only 13 placeholder zeros, indicating that the actual measurement value is 840,000,000,000,000. Numbers that are fractions can be indicated by scientific notation as well. Consider the number 0.0000045. Its scientific notation is 4.5 × 10–6. Its scientific notation has the same format x × 10y Here, x is 4.5. However, the value of y in the 10y factor is negative, which indicates that the measurement is a fraction of 1. Therefore, we move the decimal place to the left, for a negative y. In our example of 4.5 × 10–6, the decimal point would be moved to the left six times to yield the original number, which would be 0.0000045. UNDERSTANDING: Post-test Activity 1 Convert the given quantities: 1. 2. 3. 4. 5. 6. 150 cm to m 360 mm to m 2100 cm3 to l 1.2 GV to V 4.6 ms to s 450 K to 0F 6 Specialized Subject - STEM Activity 2 • Express the following numbers in scientific notation. 1. 98 2. 0.0026 3. 0.0000401 4. 643.9 5. 816 6. 45800 7. 0.0068 8. 5600 9. 902 10. 0.0045 • Transform the following scientific notation to standard notation 1. 6.455 x 104 2. 3.1 x 10-6 3. 5.00 x 10-2 4. 7.2 x 103 5. 9 x 105 6. 7.4 x 10-3 7. 9.3 x 102 8. 2.5 x 10-4 9. 4.01 x 101 10. 2.4 x 100 Post Test A. Answer the following problems. Show your solution. 1. A certain box has the following dimensions: 20 in x 7 cm x 200 mm. Calculate the volume of the box in cubic meter. 2. Find the volume of sphere (in millimeter) with a radius of 25 inches. 7 Specialized Subject - STEM 3. John weighs 320 pounds. His brother weighs 20,500 grams. What is their weight difference in kilogram? 4. A bond paper measured to be 320 mm wide. What is its width in cm? In meters? 5. If one mile is equivalent to 5280 ft long. How many cm are there in a mile? How many km? 6. Old laboratory manual uses 7.5 kg of sodium chloride to perform the experiment, but no available sodium chloride in kg instead the container of the chemical is express in grams. How many grams of sodium chloride will you use if you wish to continue the experiment? 7. Peter uses his hand to measure the width of his room. If his hand has a width of 6 in and he finds the room to be 20 hand wide, what is the width of the room in cm? In meter? 8. The water tank in a St. Jude subdivision holds 10 kiloliters of water. How many liters is this? How many mL? 9. According to the expert the seafloor is expanding by about 5 cm a year. What will be the seafloor’s expansion in meters after 5 years? After 6 ½ years? 10. The area of a rectangle is 550 cm3. If its width is 300mm, what is its length in meters? 8 Specialized Subject - STEM B. Write the answer for each expression using scientific notation with the appropriate number of significant figures. 1. 23.096 × 90.3 = 2. 125 × 9.00 = 3. 1.67 𝑥 1027 = 2.5 𝑥 102 5 𝑥 10 4. +1.67 4.6 𝑥 102 = 5. (4.5 x 10-14) x (5.2 x 103) = 5 𝑥 10 6. −4.5 3.55 𝑥 104 = 7. (3.74 x 10-3)4 = 8. 9. 6.1 𝑥 105 1.2 𝑥 10−3 5.0 𝑥 10−3 2.5 𝑥 10−6 = = 6 𝑥 10 10. −7.85 6.7 𝑥 102 = C. Write in scientific notation: 11. 0.000467 = 12. 32,000,000 = 13. 0.000 000 245 = 14. 567,000,000,000 = 15. 0.000,000,950 = D. Express the following as a number 16. 5.43 x 10-3 = 9 Specialized Subject - STEM 17. 3.70 x 106 = 18. 1.59 x 10-5 = 19. 4.1 x 104 = 20. 1.4 x 10-3 = 10 Specialized Subject - STEM MODULE1Q1 MEASUREMENTS Lesson 2: Accuracy and Precision Overview: This module was designed and written with you in mind. It is here to help you master the accuracy and precision. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Differentiate accuracy from precision Lesson Learning Objectives: After going through this module, you are expected to: ✓ define accuracy and precision; ✓ differentiate accuracy and precision; and ✓ illustrate an example of accuracy and precision PRE-TEST Direction: Read the questions properly, choose the correct answer and write it on the blanks provided before each item. ____1. It refers to the degree to which successive measurements agree with each other. a. accuracy c. reliability b. precision d. validity ____2. It is described as the degree of how close the measurements are to the true value. a. accuracy c. reliability 11 Specialized Subject - STEM b. precision d. validity ____3. What is meant by the term accuracy? a. The extent to which the value approaches its true value. b. The level of detail at which data is stored. c. The lack of bias in the data. d. The overall quality of data ____4. What is meant by the term precision? a. The extent to which the value approaches its true value. b. The level of detail at which data is stored. c. The lack of bias in the data. d. The overall quality of data _____5. Which of the following will allow measurement of a liquid's volume with the greatest precision? a. 50 ml cylinder graduated in 1ml increments b. 100 ml cylinder graduated in 0.5 ml increments c. 100 ml cylinder graduated in 1 ml increments d. 200 ml cylinder graduated in 5 ml increments 12 Specialized Subject - STEM PRIMING Accuracy vs Precision PROCESSING Accuracy It is how close a measurement is to the correct value for that measurement. For example, let us say that you are measuring the length of standard piece of bond paper. The packaging in which you purchased the paper states that it is 11 inches long, and suppose this stated value is correct. You measure the length of the paper three times and obtain the following measurements: 11.1 inches, 11.2 inches, and 10.9 inches. These measurements are quite accurate because they are very close to the correct value of 11.0 inches. In contrast, if you had obtained a measurement of 12 inches, your measurement would not be very accurate. This is why measuring instruments are calibrated based on a known measurement. If the instrument consistently returns the correct value of the known measurement, it is safe for use in finding unknown values. Precision It states how well repeated measurements of something generate the same or similar results. Therefore, the precision of measurements refers to how close together the measurements are when you measure the same thing several times. One way to analyze the precision of measurements would be to determine the range, or difference between the lowest and the highest measured values. In the case of the printer paper Specialized Subject - STEM measurements, the lowest value was 10.9 inches and the highest value was 11.2 inches. Thus, the measured values deviated from each other by, at most, 0.3 inches. These measurements were reasonably precise because they varied by only a fraction of an inch. However, if the measured values had been 10.9 inches, 11.1 inches, and 11.9 inches, then the measurements would not be very precise because there is a lot of variation from one measurement to another. The measurements in the paper example are both accurate and precise, but in some cases, measurements are accurate but not precise, or they are precise but not accurate. Let us consider a GPS system that is attempting to locate the position of a restaurant in a city. Think of the restaurant location as existing at the center of a bull’s-eye target. Then think of each GPS attempt to locate the restaurant as a black dot on the bull’s eye. UNDERSTANDING: Post-test Activity 1 Direction: Answer the following. A. Five groups of students measured the same volume of water. They all made three trials and tabulated their results in the table below. Suppose that the accepted value is 20.50 mL, how would you describe their measurements in terms of accuracy and precision? Trial Group 1 Group 2 Group 3 Group 4 Group 5 1 17.77 mL 23.93 mL 20.50 mL 20.50 mL 19.60 mL 2 19.65 mL 23.92 mL 20.45 mL 20.49 mL 19,55 mL 3 25.15 mL 23.94 mL 20.55 mL 20.51 mL 19.75 mL Σ 20.86 mL 23.93 mL 20.50 mL 20.49 mL 19.63 mL 1. Which group has high precision? ___2__, __3___, __3___, ___3__ 2. Which group has high accuracy? __3___, and ___4__ 3. Which group has high precision but low accuracy? ____2___ and ___5___ 4. Which group has high accuracy and low precision? ___1____ 5. Explain the statement “High precision does not necessarily mean high accuracy.” ___________________________________________________________________________ ___________________________________________________________________________ ____________________________________. 6. Jared is practicing for a golf tournament. His normal driver distance is 250 yards. He hits three balls with his driver, and they travel a distance of 190 yards, 195 yards, and 187 yards. Which of the following is true? Specialized Subject - STEM a. b. c. d. His drives are accurate but not precise. His drives are precise but not accurate. His drives are both accurate and precise. His drives are neither accurate nor precise. Activity 2 B. Refer to the following experimental data of the three groups of students working in the laboratory. Trials Group A Group B Group C 1 0.980 g/mL 0.732 g/mL 1.023 g/mL 2 1.020 g/mL 0.731 g/mL 0.739 g/mL 3 0.970 g/mL 0.733 g/mL 0.845 g/mL True Value: 1.000 g/mL 7. Which group has high accuracy? ______ Why? Explain. ______________________________________________________________ ______________________________________________________________ 8. Which group has high precision? __________ Why? Explain. ______________________________________________________________ ______________________________________________________________ 9. Which group shows high precision but has poor accuracy? __________ Explain. ___________________________________________________________________ ___________________________________________________________________ C. The density of mercury is 13.55 g/cm3. Experimental results give the following data: 10.45/cm3 16.56g/cm3 15.75g/cm3 12.35g/cm3 10. Can the measurements be described as accurate? Precise? Post Test Multiple Choice. Encircle the letter of the best answer. 1. It is described as the degree of how close the measurements are to the true value. a. accuracy c. reliability b. precision d. validity Specialized Subject - STEM 2. It refers to the degree to which successive measurements agree with each other. a. accuracy c. reliability b. precision d. validity 3. Which group of measurements is most precise? a. 0.005 g, 0.0049 g, 0.0051 g b. 1.23 cm3, 2.21 cm3, 9.92 cm3 c. 23.4 mm, 12.4 mm, 50.2 mm d. 2.3 x 10-2 kg, 2.31 x 102 kg, 2.29 x 1012 kg 4. The volume of a liquid is 20.5 ml. Which of the following sets of measurement the value with good accuracy? a. 18.6 ml, 17.6 ml, 19.6 ml, 17.2 ml b. 18.8 ml, 19.0 ml, 19.2 ml, 18.8 ml. c. 19.3 ml, 19.2 ml, 18.6 ml, 18.7 ml d. 20.2 ml, 20.5 ml, 20.3 ml 20.1 ml 5. The mass of unknown substance is 2.86 g. Which of the following sets of measurement represents the value with both accuracy and precision? a. 1.78 g, 1.80 g, 1.76 g, 1.81 g b. 1.95 g, 2.02 g, 1.96 g, 2.01 g c. 2.81 g, 1.98 g, 2.40 g, 2.78 g d. 2.85 g, 2.86 g, 2.84 g, 2.81 g 6. The mass of a sample of a copper nitrate is 3.82 g. A student measures the mass and finds it to be 3.81 g, 3.82 g, 3.79 g and 3.80 g in the first, second, third and fourth trial, respectively. Which of the following statements is true for his measurements? a. They have good accuracy but poor precision. b. They have poor accuracy but good precision. c. They are neither precise nor accurate. Specialized Subject - STEM d. They have good accuracy and precision. Mass Data Sample TRIAL 1 TRIAL 2 TRIAL 3 TRIAL 4 Student A 1.43 g 1.52 g 1.47 g 1.42 g Student B 1.43 g 1.40 g 1.46 g 1.44 g Student C 1.54 g 1.56 g 1.58 g 1.50 g Student D 0.86 g 1.24 g 1.52 g 1.42 g 7. Four students each measured the mass of one 1.43 g sample four times. The results in the data above indicate that the data collected by reflect the greatest accuracy and precision. a. Student A b. Student B c. Student C d. Student D 8. The accepted value is 1.43. Which correctly describes this student’s experimental data? Trial Measurement 1 1.29 2 1.93 3 0.88 a. Accurate but not precise c. Precise but not accurate b. Both accurate and precise d. Neither accurate nor precise 9. What is meant by the term accuracy? a. The extent to which the value approaches its true value. b. The level of detail at which data is stored. Specialized Subject - STEM c. The lack of bias in the data. d. The overall quality of data. 10. What is meant by the term precision? a. The extent to which the value approaches its true value. b. The level of detail at which data is stored. c. The lack of bias in the data. d. The overall quality of data. 11. The volume of a liquid is 25.5 ml. A student measures the volume and finds it to be 25.2 mL, 25.1 mL, 24.9 mL, and 25.3 mL in the first, second, third, and fourth trial, respectively. Which of the following statements is true for his measurements? a. They have poor precision. b. They have poor accuracy. c. They are neither precise nor accurate. d. They have good precision. 12. The mass of an unknown substance is 2.86 g. Which of the following sets of measurement represents the value with both accuracy and precision? a. 1.78 g, 1.80 g, 1.76 g, 1.81 g b. 1.98 g, 2.02 g, 1.96 g, 2.01 g c. 2.85 g, 2.86 g, 2.84 g, 2.81 g d. 2.81 g, 1.98 g, 2.40 g, 2.78 g 13. The volume of a sample of concentrated hydrochloric acid is 10.5 ml. A student measures the volume and finds it to be 8.6 mL, 8.8 mL, 8.2 mL, and 8.6 mL in the first, second, third, and fourth trial, respectively. Which of the following statements is true for his measurements? a. They have poor precision. b. They have poor accuracy. c. They are neither precise nor accurate. d. They have good precision. Specialized Subject - STEM 14. Looking at the above rifle target, how would you describe the shooting of this contestant? a. accurate and imprecise b. accurate and precise c. inaccurate and precise d. inaccurate and imprecise 15. Which of the following will allow measurement of a liquid's volume with the greatest precision? a. 50 ml cylinder graduated in 1ml increments b. 100 ml cylinder graduated in 0.5 ml increments c. 100 ml cylinder graduated in 1 ml increments d. 200 ml cylinder graduated in 5 ml increments Specialized Subject - STEM MODULE1Q1 MEASUREMENTS Lesson 3: Random Error and Systematic Error Overview: This module was designed and written with you in mind. It is here to help you master the random error and systematic error. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Differentiate random errors from systematic errors Lesson Learning Objectives: After going through this module, you are expected to: ✓ define random and systematic error; ✓ differentiate random and systematic error; and ✓ illustrate an example of random and systematic error PRE-TEST Direction: Read the questions properly, choose the correct answer and write it on the blanks provided before each item. ____1. In measuring the diameter circular object like coins using Vernier caliper may reduce what kind of error? a. neither random nor systematic error b. random error c. random and systematic error d. systematic error ____2. Random error lead to a lack of: a. accuracy in measurement b. gradation of measuring instrument Specialized Subject - STEM c. precision in measurement d. significant digits in measurement ____3. Which of these is not true for systematic errors? a. They arise due to errors in the measuring instrument used. b. They are reproducible that are consistently in the same direction. c. Repeating the observations or increasing the sample size can eliminate them. d. They arise from the design of the study. ____4. A group of measurements for which there is insignificant random error but significant systematic error is a. imprecise and biased b. imprecise and unbiased c. precise and biased d. precise and unbiased _____5 Which of these is not true for random errors? a. They are difficult to detect. b. They are less likely for small sizes. c. They do not arise from the design of the study PRIMING Random errors It is usually result from the experimenter’s inability to take the same measurement in exactly the same way to get exact the same number. Systematic errors There are reproducible inaccuracies that are consistently in the same direction. Systematic errors are often due to a problem which persists throughout the entire experiment. Note that systematic and random errors refer to problems associated with making measurements. Mistakes made in the calculations or in reading the instrument are not considered in error analysis. It is assumed that the experimenters are careful and competent! Specialized Subject - STEM PROCESSING Definition of Random Error The uncertain disturbances occur in the experiment is known as the random errors. Such type of errors remains in the experiment even after the removal of the systematic error. The magnitude of error varies from one reading to another. The random errors are inconsistent and occur in both the directions. The presence of random errors is determined only when the different readings are obtained for the measurement of the same quantity under the same condition. Definition of Systematic Error The constant error occurs in the experiment because of the imperfection of the mechanical structure of the apparatus is known as the systematic error. The systematic errors arise because of the incorrect calibration of the device. The error is mainly categorized into three types. • Instrumental Error • Environmental Error • Observational Error Specialized Subject - STEM Instrumental Error – The instrumental error occurs because of the three reasons. 1. Misuse of the apparatus. 2. Imperfection in the mechanical structure of the apparatus. 3. The error occurs because of the loading effect. UNDERSTANDING: Post-test Activity 1 A. Classify the statement below as : Systematic Error or Random Error 1. When weighing yourself on a scale, you position yourself slightly differently each time. _______________ 2. Forgetting to tare or zero a balance produces mass measurements that are always "off" by the same amount. An error caused by not setting an instrument to zero prior to its use is called an offset error. _______________ 3. When taking a volume reading in a flask, you may read the value from a different angle each time. _______________ 4. Not reading the meniscus at eye level for a volume measurement will always result in an inaccurate reading. The value will be consistently low or high, depending on whether the reading is taken from above or below the mark. _______________ 5. Measuring the mass of a sample on an analytical balance may produce different values as air currents affect the balance or as water enters and leaves the specimen. _______________ 6. Measuring length with a metal ruler will give a different result at a cold temperature than at a hot temperature, due to thermal expansion of the material. _______________ 7. Measuring your height is affected by minor posture changes. _______________ 8. An improperly calibrated thermometer may give accurate readings within a certain temperature range, but become inaccurate at higher or lower temperatures. _______________ 9. Measuring wind velocity depends on the height and time at which a measurement is taken. Multiple readings must be taken and averaged because gusts and changes in direction affect the value. _______________ 10. Measured distance is different using a new cloth measuring tape versus an older, stretched one. Proportional errors of this type are called scale factor errors. _______________ 11. Readings must be estimated when they fall between marks on a scale or when the thickness of a measurement marking is taken into account. Specialized Subject - STEM _______________ 12. Drift occurs when successive readings become consistently lower or higher over time. Electronic equipment tends to be susceptible to drift. Many other instruments are affected by (usually positive) drift, as the device warms up. ________________ Post Test Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 1. In a zoology class, Pedro measured the length of an earthworm using a ruler for three times as part of the data gathering procedure of his experiment. What type of measurement error might he commit? a. calibration error c. random error b. human error d. systematic error 2. April forgot to calibrate her analytical balance before she measured the mass of her reactants in a chemistry experiment. She committed 78% percentage error in her measurement. What type of measurement error did she commit? a. human error c. random error b. parallax error d. systematic error 3. The observation error of a measured quantity a. corresponds to the random error in the measurement b. the difference between the measured and true values and is inevitably present c. the result of a mistake or blunder but can be reduced by taking several measurements and averaging them 4. A group of measurements for which there is insignificant random error but significant systematic error is a. imprecise and biased c. precise and biased b. imprecise and unbiased d. precise and unbiased Specialized Subject - STEM 5. Compared to the precision of individual measurements, the arithmetic mean of 150 measurements subject to random error can be written using a. one additional significant digit b. one fewer significant digit c. two additional significant digits 6. Which of these is not true for systematic errors? a. They arise due to errors in the measuring instrument used. b. They are reproducible that are consistently in the same direction. c. Repeating the observations or increasing the sample size can eliminate them. d. They arise from the design of the study. 7. Which of these is not true for random errors? a. They are difficult to detect. b. They are less likely for small sizes. c. They do not arise from the design of the study. 8. Systematic errors lead to a lack of: a. accuracy in measurement b. gradation of measuring instrument c. precision in measurement d. significant digits in measurement 9. Random error lead to a lack of: a. accuracy in measurement Specialized Subject - STEM b. gradation of measuring instrument c. precision in measurement d. significant digits in measurement 10. Repeated measurement of quantity can reduce the effects of a. both random and systematic errors b. neither random errors nor systematic errors c. random errors d. systematic errors 11. Which of the following statements is INCORRECT regarding systematic error? a. It is the same as random error b. it can be minimized by increasing the study samples. c. it can be increased by increasing the study samples. d. it occurs as a result of “the luck of the draw” an inaccurate estimate resulting from the sample that was not representative of the population. 12. Which of the following statements is true regarding systematic error? a. It is the same as random error b. it can be minimized by increasing the study samples. c. it can be increased by increasing the study samples. d. it occurs as a result of “the luck of the draw” an inaccurate estimate resulting from the sample that was not representative of the population. 13. In measuring the diameter circular object like coins using Vernier caliper may reduce what kind of error? a. neither random nor systematic error Specialized Subject - STEM b. random error c. random and systematic error d. systematic error 14. To check the exact mass of set of weights 1kg you use the triple beam balance you need to calibrate this measuring device, what kind of error did you try to minimize? a. neither random nor systematic error b. random error c. random and systematic error d. systematic error 15. In using the multi-tester to measure the resistance value of the ohmic material you need to calibrate the device, what kind of error do want to decreased the value? a. neither random nor systematic error b. random error c. random and systematic error d. systematic error Specialized Subject - STEM MODULE1Q1 MEASUREMENTS Lesson 4: Least Concept to Estimate Error Overview: This module was designed and written with you in mind. It is here to help you master the measurements. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Estimate errors from multiple measurements of a physical quantity using variance Lesson Learning Objectives: After going through this module, you are expected to: ✓ Use the least count concept to estimate errors associated with single measurements. PRE-TEST Direction: Read the questions properly, choose the correct answer and write it on the blanks provided before each item. ____1. The following observations have been made: 64.52, 3.0, 11.081. the correctly written sum is a. 78.6 b. 78.60 c. 78.6010 ____2. Which of the following numbers contains the designated CORRECT number of significant figures? a. 0.00302 2 significant figures b. 0.04300 5 significant figures c. 1.04 2 significant figures d. 3.0560 4 significant figures Specialized Subject - STEM e. 156 000 3 significant figures ____3. How many significant figures are in the measurement of 102 400 meters? a. three b. four c. five d. six ____4. The sum of 1.04 + 2.1135 + 3.1 + 3.403 is_____ a. 9.6565 c. 9.66 b. 9.6 d. 9.70 _____5. The mass of a watch glass was measured four times. The masses were 99.997 g, 100.008 g, 100.001 g, 100.005 g. What is the average mass of the watch glass? a. 100.00 g c. 100.005 g b. 100.01 g d. 100.00525 g PRIMING To determine the number of significant figures in a number use the following 3 rules: 1. Non-zero digits are always significant 2. Any zeros between two significant digits are significant 3. A final zero or trailing zeros in the decimal portion ONLY are significant Example: .500 or .632000 the zeros are significant .006 or .000968 the zeros are NOT significant For addition and subtraction use the following rules: 1. Count the number of significant figures in the decimal portion ONLY of each number in the problem 2. Add or subtract in the normal fashion 3. Your final answer may have no more significant figures to the right of the decimal than the LEAST number of significant figures in any number in the problem. For multiplication and division use the following rule: Specialized Subject - STEM 1. The LEAST number of significant figures in any number of the problem determines the number of significant figures in the answer. (You are now looking at the entire number, not just the decimal portion) *This means you have to be able to recognize significant figures in order to use this rule* Example: 5.26 has 3 significant figures 6.1 has 2 significant figures PROCESSING Rules for Significant Figure 1. All non-zero numbers ARE significant. The number 33.2 has THREE significant figures because all of the digits present are non-zero. 2. Zeros between two non-zero digits ARE significant. 2051 has FOUR significant figures. The zero is between a 2 and a 5. 3. Leading zeros are NOT significant. They're nothing more than "place holders." The number 0.54 has only TWO significant figures. 0.0032 also has TWO significant figures. All of the zeros are leading. 4. Trailing zeros to the right of the decimal ARE significant. There are FOUR significant figures in 92.00. 92.00 is different from 92: a scientist who measures 92.00 milliliters knows his value to the nearest 1/100th milliliter; meanwhile his colleague who measured 92 milliliters only knows his value to the nearest 1 milliliter. It's important to understand that "zero" does not mean "nothing." Zero denotes actual information, just like any other number. You cannot tag on zeros that aren't certain to belong there. 5. Trailing zeros in a whole number with the decimal shown ARE significant. Placing a decimal at the end of a number is usually not done. By convention, however, this decimal indicates a significant zero. For example, "540." indicates that the trailing zero IS significant; there are THREE significant figures in this value. 30 Specialized Subject - STEM 6. Trailing zeros in a whole number with no decimal shown are NOT significant. Writing just "540" indicates that the zero is NOT significant, and there are only TWO significant figures in this value. 7. Exact numbers have an INFINITE number of significant figures. This rule applies to numbers that are definitions. For example, 1 meter = 1.00 meters = 1.0000 meters = 1.0000000000000000000 meters, etc. So now back to the example posed in the Rounding Tutorial: Round 1000.3 to four significant figures. 1000.3 has five significant figures (the zeros are between non-zero digits 1 and 3, so by rule 2 above, they are significant.) We need to drop the final 3, and since 3 < 5, we leave the last zero alone. so 1000. is our four-significant-figure answer. (from rules 5 and 6, we see that in order for the trailing zeros to "count" as significant, they must be followed by a decimal. Writing just "1000" would give us only one significant figure.) 8. For a number in scientific notation: N x 10x, all digits comprising N ARE significant by the first 6 rules; "10" and "x" are NOT significant. 5.02 x 104 has THREE significant figures: "5.02." "10 and "4" are not significant. Rule 8 provides the opportunity to change the number of significant figures in a value by manipulating its form. For example, let's try writing 1100 with THREE significant figures. By rule 6, 1100 has TWO significant figures; its two trailing zeros are not significant. If we add a decimal to the end, we have 1100., with FOUR significant figures (by rule 5.) But by writing it in scientific notation: 1.10 x 103, we create a THREE-significant-figure value. UNDERSTANDING: Post-test Activity A. Identify the numbers of significant figures of the following: 1. 4308 – 2. 40.05 – 3. 470,000 – 4. 4.00 – 5. 0.00500 – 31 Specialized Subject - STEM B. Round the following into 1 significant figures, then into 2 significant figures 6. 53,879 to 1 significant figure, then 2 significant figures. 1 significant figure is __________ 2 significant figures is _________ 7. Round 0.005089 to 1 significant figure, then 2 significant figures. 1 significant figure is __________ 2 significant figures is __________ 8. What is 98,347 rounded to 1 significant figure, then 2 significant figures? 1 significant figure is __________ 2 significant figures is __________ 9. What is 3.5175 rounded to 1 significant figure, then 2 significant figures 1 significant figure is __________ 2 significant figures is __________ C. Complete the table: Rounding the number 205.469 Rounded to Rounded to Rounded to how many SF Significant Decimal or DP Figures (SF) Places (DP) 10. 0 11. 1 12. 2 13. 3 14. 4 15. 5 16. 6 - D. Answer the following. Express your answer based from the number of significant figures. 17. Find the product of 47.5 x 0.52 x 1.5 = __________ 32 Specialized Subject - STEM 18. If the object has a mass of 10.00 g and has a volume of 5.5mL. If the formula for finding 𝑚 the density is = 𝑉 , what is the density of the said solid? 19. The volume of a cylinder is given by 𝑉 = 𝜋𝑟 2 ℎ. Find the volume of a cylinder with a height of 659.75 cm and a radius of 20.015 cm. 20. John weighs 64.8 kg. His brother weighs 76 kg. How much heavier is his brother? Post Test Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 1. Considering the correct number of significant figures, evaluate the following operation, 3.73 x 5.7 = _____. a. 21 c. 21.26 b. 21.00 d. 21.261 2. Compute 3.24 m + 0.532 m to the correct number of significant figures. a. 3.77 c. 3.8 b. 3.772 d. 4.00 3. The sum of 1.04 + 2.1135 + 3.1 + 3.403 is_____ a. 9.6565 c. 9.66 b. 9.6 d. 9.70 4. Solve: 7.45 x 108 + 4.97 x 10-2 – 6.67 x 105 is equal to___ a. 7443.33 x 105 b. 7.44 x 108 c. 7.44333 x 10-2 d. 7443.330000497 33 Specialized Subject - STEM 5. Which of the following examples illustrates a number that is correctly rounded to three significant figures? a. 0.03954 g to 4.040 g c. 20.0332 g to 20.0 g b. 4.05438 g to 4.054 g d. 103.692 g to 103.7 g 6. Which of the following numbers contains the designated CORRECT number of significant figures? a. 0.00302 2 significant figures b. 0.04300 5 significant figures c. 1.04 2 significant figures d. 3.0560 4 significant figures e. 156 000 3 significant figures 7. A calculator answer of 423.6059 must be rounded off to three significant figures. What answer is reported? a. 420 b. 423 c. 423.6 d. 423.7 e. 424 8. Which of the following is CORRECT? a. 2.450 x 107 rounded to two significant digits 2.4 x 107 b. 3.56 rounded to two significant digits is 3.6 c. 77.889 x 106 rounded to three significant digits is 77.8 x 106 d. 122.5 rounded to two significant digits is 120 34 Specialized Subject - STEM 9. The following observations have been made: 64.52, 3.0, 11.081. the correctly written sum is a. 78.6 b. 78.60 c. 78.6010 d. 79 10. The quantity 0.245 x 36.74 / 200.0 = 0.045007, computed from measured values, should be written in an engineering report as c. 4.50 x 10-2 a. 0.04500 b. 4.5 x 10-2 d. 5 x 10-2 11. The mass of a watch glass was measured four times. The masses were 99.997 g, 100.008 g, 100.001 g, 100.005 g. What is the average mass of the watch glass? a. 100.00 g c. 100.005 g b. 100.01 g d. 100.00525 g 12. When performing the calculation 34.530 g + 12.1 g + 1 222.34 g, the final answer must have a. only one decimal place c. three significant figures d. unit of g3 b. three decimal places 13. How many significant figures are in the measurement of 102 400 meters? a. three b. four c. five d. six 14. 923 g is divided by 20 312 cm3 a. 0.045 g/cm3 c. 0.0454 g/cm3 b. 4.00 x 10-2 g/cm3 d. 0.04 g/cm3 35 Specialized Subject - STEM 15. Complete the following problem: A piece of stone has a mass of 24.595 grams and a volume of 5.34 cm3. What is the density of the stone? (remember that density = m/v) a. 0.22 cm3/g b. 4.606 g/cm3 c. 4.61 g/cm3 d. 0.217 cm3/g 36 Specialized Subject - STEM MODULE1Q1 MEASUREMENTS Lesson 5: Estimate error using variance Overview: This module was designed and written with you in mind. It is here to help you master the estimate error using variances. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Estimate errors from multiple measurements of a physical quantity using variance Lesson Learning Objectives: After going through this module, you are expected to: ✓ Estimate errors from multiple measurements of a physical quantity using variance; and ✓ Calculate standard deviation and percentage error PRE-TEST Direction: Read the questions properly, choose the correct answer and write it on the blanks provided before each item. ____1. Olivia measured the length and width of a rectangular garden, each to the nearest 0.1 yd. She recorded the length of the garden as 41.5 yds. and the width of the garden as 30.8 yds. Which of the following is true for the area A yds2 of the garden? a. 1274.5875 A 1281.75 b. 1278.15 A 1278.25 c. 1274.5875 A 1281.75 d. 1278.2 2. Garth wanted to find the area of a square. He measured the length of the square as 2 cm. Later, the actual length of the square was more accurately measured as 2.1 cm. What is the relative error in his area calculation to the nearest hundredth? a. .01 c. .09 37 Specialized Subject - STEM b. .08 d. 0.10 3. Kyle wanted to find the area of a circle. He measured the radius of the circle as 5.4 cm. Later, the actual radius of the circle was more accurately measured as 5.35 cm. What is the relative error in his area calculation to the nearest thousandth? a. .018 c. .020 b. .019 d. .022 4. In an experiment, the temperature of a solution is measured by a student to be 79 degrees, but the true value of the temperature is 85 degrees. What is the percent error in this measurement? a. .07% b. 1.07% c. 7.1% d. 92% 5. A student measured the length of a table to be 65 cm, but the table was actually 62 cm long. What was the percent error in this measurement? a. 0.95% b. 1.04% c. 4.8% e. 48% PRIMING Absolute, Relative and Percentage Error The Absolute Error is the difference between the actual and measured value. But ... when measuring we don't know the actual value! So we use the maximum possible error. In the example above the Absolute Error is 0.05 m What happened to the ± ...? Well, we just want the size (the absolute value) of the difference. The Relative Error is the Absolute Error divided by the actual measurement. We don't know the actual measurement, so the best we can do is use the measured value: Relative Error = Absolute Error Measured Value 38 Specialized Subject - STEM The Percentage Error is the Relative Error shown as a percentage. PROCESSING Example: a fence is measured as 12.5 meters long, accurate to 0.1 of a meter Length = 12.5 ±0.05 m So: Absolute Error = 0.05 m And: Relative Error = 0.05 m 12.5 m = 0.004 And: Percentage Error = 0.4% Example: The thermometer measures to the nearest 2 degrees. The temperature was measured as 38° C The temperature could be up to 1° either side of 38° (i.e. between 37° and 39°) Temperature = 38 ±1° So: Absolute Error = 1° And: Relative Error = 1°38° = 0.0263... 39 Specialized Subject - STEM And: Percentage Error = 2.63...% Estimating Uncertainty in Repeated Measurements Suppose you time the period of oscillation of a pendulum using a digital instrument (that you assume is measuring accurately) and find: T = 0.44 seconds. This single measurement of the period suggests a precision of ±0.005 s, but this instrument precision may not give a complete sense of the uncertainty. If you repeat the measurement several times and examine the variation among the measured values, you can get a better idea of the uncertainty in the period. For example, here are the results of 5 measurements, in seconds: 0.46, 0.44, 0.45, 0.44, 0.41. (5) x1 + x2 + + xN Average (mean) = N For this situation, the best estimate of the period is the average, or mean. Whenever possible, repeat a measurement several times and average the results. This average is generally the best estimate of the "true" value (unless the data set is skewed by one or more outliers which should be examined to determine if they are bad data points that should be omitted from the average or valid measurements that require further investigation). Generally, the more repetitions you make of a measurement, the better this estimate will be, but be careful to avoid wasting time taking more measurements than is necessary for the precision required. Consider, as another example, the measurement of the width of a piece of paper using a meter stick. Being careful to keep the meter stick parallel to the edge of the paper (to avoid a systematic error which would cause the measured value to be consistently higher than the correct value), the width of the paper is measured at a number of points on the sheet, and the values obtained are entered in a data table. Note that the last digit is only a rough estimate, since it is difficult to read a meter stick to the nearest tenth of a millimeter (0.01 cm). 40 Specialized Subject - STEM Average sum of observed widths 155.96 cm no. of observations 5 = 31.19 cm This average is the best available estimate of the width of the piece of paper, but it is certainly not exact. We would have to average an infinite number of measurements to approach the true mean value, and even then, we are not guaranteed that the mean value is accurate because there is still some systematic error from the measuring tool, which can never be calibrated perfectly. So how do we express the uncertainty in our average value? One way to express the variation among the measurements is to use the average deviation. This statistic tells us on average (with 50% confidence) how much the individual measurements vary from the mean. |x1 − x| + |x2 − x| + d= + |xN − x| N However, the standard deviation is the most common way to characterize the spread of a data set. The standard deviation is always slightly greater than the average deviation, and is used because of its association with the normal distribution that is frequently encountered in statistical analyses. STANDARD DEVIATION To calculate the standard deviation for a sample of N measurement: 1 Sum all the measurements and divide by N to get the average, or mean. 2 Now, subtract this average from each of the N measurements to obtain N "deviations". 3. Square each of these N deviations and add them all up. 4 Divide this result by (N − 1) and take the square root. We can write out the formula for the standard deviation as follows. Let the N measurements be called x1, x2, ..., xN. Let the average of the N values be called x. Then each deviation is given by δxi = xi − x, for i = 1, 2, , N. The standard deviation is: 41 Specialized Subject - STEM s= (δx12 + δx22 + + δxN2) (N − 1) In our previous example, the average width x is 31.19 cm. The deviations are: The average deviation is: d = 0.086 cm. The standard deviation is: s= (0.14)2 + (0.04)2 + (0.07)2 + (0.17)2 + (0.01)2 5−1 = 0.12 cm. The significance of the standard deviation is this: if you now make one more measurement using the same meter stick, you can reasonably expect (with about 68% confidence) that the new measurement will be within 0.12 cm of the estimated average of 31.19 cm. In fact, it is reasonable to use the standard deviation as the uncertainty associated with this single new measurement. However, the uncertainty of the average value is the standard deviation of the mean, which is always less than the standard deviation (see next section). Consider an example where 100 measurements of a quantity were made. The average or mean value was 10.5 and the standard deviation was s = 1.83. The figure below is a histogram of the 100 measurements, which shows how often a certain range of values was measured. For example, in 20 of the measurements, the value was in the range 9.5 to 10.5, and most of the readings were close to the mean value of 10.5. The standard deviation s for this set of measurements is roughly how far from the average value most of the readings fell. For a large enough sample, approximately 68% of the readings will be within one standard deviation of the mean value, 95% of the readings will be in the interval x ± 2 s, and nearly all (99.7%) of readings will lie within 3 standard deviations from the mean. The smooth curve superimposed on the histogram is the gaussian or normal distribution predicted by theory for measurements involving random errors. As more and more measurements are made, the histogram will more closely follow the bell-shaped gaussian curve, but the standard deviation of the distribution will remain approximately the same. 42 Specialized Subject - STEM UNDERSTANDING: Post-test Calculate the average and standard deviation of the given width of paper. Enter your calculated deviation per observation on Column 3. Observation Width (cm) 1 31.33 2 31.15 3 31.26 4 31.02 5 31.20 Deviations (cm) Average: ___________ Standard Deviation: ______________ Computation: Interpretation: Post Test Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 1. Evaluate the percentage error of the following measurement: True Value = 89.49 km 43 Specialized Subject - STEM Trial 1 = 85.44 km Trial 2 = 82.56 km Trial 3 = 84.49 km Trial 4 = 81.45 km a. 3.16% c. 6.71% b. 5.26% d. 7.02% 2. What is the sample standard deviation from the data given 12, 13, 29, 18, 61, 35, 21? a. 15.87 b. 17.14 c. 41.98 d. 293.67 3. If a number is added to a set that is far away from the mean how does this affect standard deviation? a. increase c. stay the same b. decrease d. both increase & decrease For numbers 4-5. The density of silver is 13.35 g/cm3. Experimental results gave the following data: 16.45 g/cm3 10.56 g/cm3 12.75 g/cm3 15.35 g/cm3 4. The experimental value is_____ g/cm3. a. 11.45 c. 13.78 b. 12.26 d. 14.16 5. The percentage error of the measurement is a. 1% c. 3% 44 Specialized Subject - STEM b. 2% d. 4% 6. Alec measured the width and height of a rectangle, but was only able to measure them to the nearest centimeter. He recorded the width as 8 cm and the height as 5 cm. Which of the following is true for the area A cm2 of the rectangle? a. 40 c. 33.75 A 46.75 b. 39.5 A 40.5 d. 33.75 A 46.75 7. Benny measured the width and height of a rectangle, but was only able to measure them to the nearest foot. He recorded the width as 12 feet and the height as 5 feet. Which of the following is true for the area A ft2 of the rectangle? a. 51.75 A 68.75 c. A = 60 b. 51.75 A 68.75 d. 59.5 A 60.5 8. Olivia measured the length and width of a rectangular garden, each to the nearest 0.1 yd. She recorded the length of the garden as 41.5 yds. and the width of the garden as 30.8 yds. Which of the following is true for the area A yds2 of the garden? a. 1274.5875 A 1281.75 c. 1274.5875 A 1281.75 b. 1278.15 A 1278.25 d. 1278.2 9. Garth wanted to find the area of a square. He measured the length of the square as 2 cm. Later, the actual length of the square was more accurately measured as 2.1 cm. What is the relative error in his area calculation to the nearest hundredth? a. .01 c. .09 b. .08 d. 0.10 10. Kyle wanted to find the area of a circle. He measured the radius of the circle as 5.4 cm. Later, the actual radius of the circle was more accurately measured as 5.35 cm. What is the relative error in his area calculation to the nearest thousandth? a. .018 b. .019 c. .020 d. .022 45 Specialized Subject - STEM 11. In an experiment, the temperature of a solution is measured by a student to be 79 degrees, but the true value of the temperature is 85 degrees. What is the percent error in this measurement? a. .07% b. 1.07% c. 7.1% d. 92% 12. A student measured the length of a table to be 65 cm, but the table was actually 62 cm long. What was the percent error in this measurement? a. 0.95% b. 1.04% c. 4.8% e. 48% 13. The period of oscillation of a simple pendulum is given by where l is about 100 cm and is known to 1mm accuracy. The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is a. 0.1% b. 0.2% c. 0.8% d. 1% 14. The percentage errors in the measurement of mass and speed are 2% and 3% respectively. How much will be the maximum error in the estimation of the kinetic energy obtained by measuring mass and speed? a. 1% b. 5% c. 8% d. 11% 15. While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His percentage error in the measurement of a. 2% b. 4% c. 7% by the relation will be d. 10% 46 Specialized Subject - STEM MODULE2Q1 VECTORS Lesson 1: Vectors and scalars Addition of Vectors Overview: This module was designed and written with you in mind. It is here to help you master the Vectors. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Differentiate vector and scalar quantities and perform addition of vectors. Lesson Learning Objectives: After going through this module, you are expected to: ✓ ✓ ✓ ✓ ✓ define scalar and vector quantity; differentiate vector and scalar quantities; classify the physical quantities as scalar and vector quantity; determine the magnitude and direction of a given vector; and perform addition of vectors PRE-TEST Direction: Read the questions properly, encircle the letter of the correct answer. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 1. Which of the following is an example of a vector quantity? a. acceleration b. mass c. volume d. temperature 2. Displacement is a a. base quantity b. derived quantity c. scalar quantity d. vector quantity 47 Specialized Subject - STEM 3. Identify the following quantities as scalar or vector: the mass of an object, the number of leaves on a tree and wind velocity. a. vector, scalar, scalar c. scalar, scalar, vector b. vector, scalar, vector d. scalar, vector, vector 4. If two forces 20 N towards North and 12 N towards South are acting on an object. What will be the resultant force? a. 32 N North b. 20 N South c. 32 N South d. 8 N North 5. A student adds two displacement vectors with magnitudes of 3 m and 4 m respectively. Which one of the following could not be a possible choice for the resultant? a. 1.3 m b. 3.3 m c. 5 m d. 6.8 m 6. Find the displacement a hiker walks if he travels 9.0 km north, and then turns around and walks 3.0 km south? a. 0.5 km c. 6.0 km b. 3.0 km d. 12.0 km 7. A runway dog walks 0.64 km due N. He then runs due W to a hot dog stand. If the magnitude of the dog’s total displacement vector is 0.91 km, what is the magnitude of the dog’s displacement vector in the due west direction? a. 0.27 km b. 0.33 km c. 0.41 km d. 0.52 km 8. An escaped convict runs 1.70 km due East of the prison. He then runs due North to a friend’s house. If the magnitude of the convict’s total displacement vector is 2.50 km, what is the direction of his total displacement vector with respect to due East? a. 340 SE b. 430 SE c. 470 NE d. 560 NE 9. Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A + B = C. Which one of the following statements concerning these vectors is true? 48 Specialized Subject - STEM a. A and B must be displacements b. A and B must have equal lengths c. A and B must point in opposite directions d. A and B point in the same direction 10. Which expression is FALSE concerning the vectors are shown in the sketch? C B A a. C=A+B b. C + A = -B c. A + B + C = 0 d. C A + B PRIMING We come into contact with many physical quantities in the natural world on a daily basis. For example, things like time, mass, weight, force, and electric charge, are physical quantities with which we are all familiar. We know that time passes and physical objects have mass. Things have weight due to gravity. We exert forces when we open doors, walk along the street and kick balls. We experience electric charge directly through static shocks in winter and through using anything which runs on electricity. There are many physical quantities in nature, and we can divide them up into two broad groups called vectors and scalars. Scalar A scalar is a physical quantity that has only a magnitude (size). For example, a person buys a tub of margarine which is labelled with a mass of 500 g. The mass of the tub of margarine is a scalar quantity. It only needs one number to describe it, in this case, 500 g. Vectors are different because they are physical quantities which have a size and a direction. A vector tells you how much of something there is and which direction it is in. 49 Specialized Subject - STEM Vector A vector is a physical quantity that has both a magnitude and a direction. For example, a car is travelling east along a freeway at 100 km/h. What we have here is a vector called the velocity. The car is moving at 100 km/h (this is the magnitude) and we know where it is going – east (this is the direction). These two quantities, the speed and direction of the car, (a magnitude and a direction) together form a vector we call velocity. Examples of scalar quantities: • mass has only a value, no direction • electric charge has only a value, no direction Examples of vector quantities: • force has a value and a direction. You push or pull something with some strength (magnitude) in a particular direction • weight has a value and a direction. Your weight is proportional to your mass (magnitude) and is always in the direction towards the center of the earth. PROCESSING Vectors are different to scalars and must have their own notation. There are many ways of writing the symbol for a vector. In this book vectors will be shown by symbols with an arrow pointing to the right above it. For example, F⃗, W⃗ and v⃗ represent the vectors of force, weight and velocity, meaning they have both a magnitude and a direction. Sometimes just the magnitude of a vector is needed. In this case, the arrow is omitted. For the case of the force vector: F⃗ represents the force vector F represents the magnitude of the force vector Graphical representation of vectors 50 Specialized Subject - STEM Vectors are drawn as arrows. An arrow has both a magnitude (how long it is) and a direction (the direction in which it points). The starting point of a vector is known as the tail and the end point is known as the head. Another common method of expressing directions is to use the points of a compass: North, South, East, and West. If a vector does not point exactly in one of the compass directions, then we use an angle. For example, we can have a vector pointing 40° North of West. Start with the vector pointing along the West direction (look at the dashed arrow below), then rotate the vector towards the north until there is a 40° angle between the vector and the West direction (the solid arrow below). The direction of this vector can also be described as: W 40° N (West 40° North); or N 50° W (North 50° West). Downloaded from vectors-and-scalars-0 https://www.siyavula.com/read/science/grade-10/vectors-and-scalars/20- Drawing vectors In order to draw a vector accurately we must represent its magnitude properly and include a reference direction in the diagram. A scale allows us to translate the length of the arrow into the vector's magnitude. For instance, if one chooses a scale of 1 cm = 2 N (1 cm represents 2 N), a force of 20 N towards the East would be represented as an arrow 10 cm long pointing towards the right. The points of a compass are often used to show direction or alternatively an arrow pointing in the reference direction. Method: Drawing Vectors 1. Decide upon a scale and write it down. 2. Decide on a reference direction 3. Determine the length of the arrow representing the vector, by using the scale. 51 Specialized Subject - STEM 4. Draw the vector as an arrow. Make sure that you fill in the arrow head. 5. Fill in the magnitude of the vector. Vector Addition Graphical techniques involve drawing accurate scale diagrams to denote individual vectors and their resultants. We will look at just one graphical method: the head-to-tail method. Method: Head-to-Tail Method of Vector Addition 1. Draw a rough sketch of the situation. 2. Choose a scale and include a reference direction. 3. Choose any of the vectors and draw it as an arrow in the correct direction and of the correct length – remember to put an arrowhead on the end to denote its direction. 4. Take the next vector and draw it as an arrow starting from the arrowhead of the first vector in the correct direction and of the correct length. 5. Continue until you have drawn each vector – each time starting from the head of the previous vector. In this way, the vectors to be added are drawn one after the other head-to-tail. 6. The resultant is then the vector drawn from the tail of the first vector to the head of the last. Its magnitude can be determined from the length of its arrow using the scale. Its direction too can be determined from the scale diagram. UNDERSTANDING: Post-test Activity 1 Categorize each quantity as being either a vector or a scalar. 1. 2. 3. 4. 5. 6. 7. 10 km 60 km/h South 40 mi downward 50 calories 250 bytes 500 m/s NE -9.8 m/s2 ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ ____________________ 52 Specialized Subject - STEM 8. 1000 kg 9. 1 hour 10. 120 m/s SW ____________________ ____________________ ____________________ Activity 2 Determine the magnitude and direction of the following vectors using a ruler and protractor. Use the scale:1 cm = 10 m/s 1. 2. 3. 53 Specialized Subject - STEM 4. Activity 3 Accurately draw scaled vector diagram to represent the magnitude and direction of the following vectors on a graphing paper. 1. 50 m 300 Scale: 1cm = 10m 2. 60 m 1500 Scale: 1cm = 10m 3. 140 m/s 2000 Scale: 1cm = 20m 4. 120 m/s 2400 Scale: 1cm = 15m/s 5. 35 m/s 2700 Scale: 1cm = 5m/s Post Test Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 54 Specialized Subject - STEM 1. Which of the following is an example of a vector quantity? a. acceleration c. volume b. mass d. temperature 2. Displacement is a a. base quantity b. derived quantity c. scalar quantity d. vector quantity 3. Identify the following quantities as scalar or vector: the mass of an object, the number of leaves on a tree and wind velocity. a. vector, scalar, scalar c. scalar, scalar, vector b. vector, scalar, vector d. scalar, vector, vector 4. If two forces 20 N towards North and 12 N towards South are acting on an object. What will be the resultant force? a. 32 N North b. 20 N South c. 32 N South d. 8 N North 5. A student adds two displacement vectors with magnitudes of 3 m and 4 m respectively. Which one of the following could not be a possible choice for the resultant? a. 1.3 m b. 3.3 m c. 5 m d. 6.8 m 6. Find the displacement a hiker walks if he travels 9.0 km north, and then turns around and walks 3.0 km south? a. 0.5 km c. 6.0 km b. 3.0 km d. 12.0 km 7. A runway dog walks 0.64 km due N. He then runs due W to a hot dog stand. If the magnitude of the dog’s total displacement vector is 0.91 km, what is the magnitude of the dog’s displacement vector in the due west direction? a. 0.27 km b. 0.33 km c. 0.41 km d. 0.52 km 8. An escaped convict runs 1.70 km due East of the prison. He then runs due North to a friend’s house. If the magnitude of the convict’s total displacement vector is 2.50 km, what is the direction of his total displacement vector with respect to due East? a. 340 SE b. 430 SE c. 470 NE d. 560 NE 55 Specialized Subject - STEM 9. Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A + B = C. Which one of the following statements concerning these vectors is true? a. A and B must be displacements b. A and B must have equal lengths c. A and B must point in opposite directions d. A and B point in the same direction 10. Which expression is FALSE concerning the vectors are shown in the sketch? C B A a. C=A+B b. C + A = -B c. A + B + C = 0 d. C A + B 11. How to add vectors graphically? a. put them in line b. put them tail to tail c. put them tip to tip d. put them tip to tail 12. Which of the following is the definition of vector? a. a quantity that has only magnitude b. a quantity that has both magnitude and direction. c. a quantity that has only one direction d. a quantity that has magnitude but may or may not have direction 13. Which of the following answer contains two scalar quantities and one vector quantity? a. mass, displacement, time c. temperature, displacement, force b. momentum, velocity, acceleration d. time, length, mass 56 Specialized Subject - STEM 14. A boy walks far 5km along a direction 530 West of North. Which of the following journeys would result in the same displacement? a. 4km N, 3 km W c. 3 km N, 2 km W b. 4 km W, 3 km W d. 3 km N, 4 km W 15. Which procedure should NOT be considered in finding the resultant vector graphically? a. use component method c. use ruler and protractor b. use head to tail method d. use scale 57 Specialized Subject - STEM MODULE2Q1 VECTORS Lesson 2: Vectors Overview: This module was designed and written with you in mind. It is here to help you master the Vectors. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Rewrite a vector in component form Lesson Learning Objectives: After going through this module, you are expected to: ✓ ✓ ✓ ✓ rewrite a vector in component form; calculate directions and magnitudes of vector; identify the x-component and y-component of the given vector; and use component method to determine the resultant vector PRE-TEST Direction: Read the questions properly, choose the correct answer encircle it. In a coordinate system, a vector is oriented at angle with respect to the x-axis. The x component of the vector equals the vector’s magnitude multiplied by which trigonometric function? a. tan b. cos 2. c. cot d. sin A particular hurricane travels 678 km, 34.60 north of west before turning into a tropical storm. Find the northern displacement of the typhoon and the western displacement of the typhoon. a. 558 km west, 385 km north b. 385 km west, 558 km north 58 Specialized Subject - STEM c. 585 km west, 358 km north d. 468 km west, 468 km north For numbers 3-4 3. Two forces act on an object. One force is 6.0 N horizontally towards west. The second force is 8.0 N vertically towards south. Find the magnitude and direction of the resultant. a. 10N 53⁰ N of E c. 10N 53⁰ E of N b. 10N 53⁰ S of W d. 10N 53⁰ W of S 4. If the object is in equilibrium, find the magnitude and direction of the force that produces equilibrium. a. 10N, 53⁰ W of S c. 10N, 53⁰ E of N b. 10N, 53⁰ N of E d. 10N, 53⁰ S of W 5. Four members of the Main Street Bicycle Club meet at a certain intersection on Main Street. The members then start from the same location but travel in different directions. A short time later, displacement vectors for the four members are: A = 2 km W B = 1.6 km N C = 2.0 km E D = 2.4 km S What is the resultant displacement R of the members of the bicycle club: R = A + B + C + D? a. 0.8 km S b. 0.4 km 450 SE c. 3.6 km 370 NW d. 4 km S 6. Given the following components for vectors A–C, find the x- and y- components for the resultant R. 59 Specialized Subject - STEM a. +11, +11 b. +7, +7 c. –7, –7 d. +7, –11 7. Given the following components for vectors A–C, find the magnitude and direction for the resultant vector R. a. b. c. d. 7, 320° in standard position 10, 40° in standard position 7, 330° in standard position 10, 30° in standard position 8. Find the x- and y-components for a displacement vector that is 23.8 km and 45.0° south of east. a. b. c. d. +16.8 km, +16.8 km –16.8 km, +16.8 km +16.8 km, –16.8 km –16.8 km, +16.8 km 9. A particular hurricane traveled 678 mi at 34.6° north of west before turning into a tropical storm. Find the northern displacement of the hurricane and the western displacement of the hurricane. a. 558 mi east, 385 mi north b. 385 mi west, 558 mi north c. 558 mi west, 358 mi north d. 468 mi west, 468 mi north 10. Find the x- and y-components to a vector that is 89.5 mm at 305° in standard position. a. –73.3 mm, 51.3 mm b. 73.3 mm, 51.3 mm 60 Specialized Subject - STEM c. –51.3 mm, 73.3 mm d. 51.3 mm, –73.3 mm 11. When resolving vectors into components or finding results __________ is/are more accurate than __________. a. geometric vector addition, geometric vector subtraction b. geometric techniques, mathematical techniques c. mathematical techniques, geometric techniques d. mathematical vector addition, mathematical vector subtraction 12. Resolve vector L into components Lx and Ly if the length of vector L is 15 m and its reference angle is 200. a. 13.9 m, 5.10 m c. 14.1, 5.13 m b. 14 m, 5 m d. 14.2, 5.20 m 13. Which is not true about vector magnitude? a. it cannot be greater than the sums of magnitude of its component vectors. b. it cannot be negative c. it is scalar quantity d. trigonometry is necessary to compute it from component vectors 14. The vector resultant of an object’s change in position is the same at its displacement. a. either true or false c. neither true nor false b. false d. true 15. Two vectors that are added together to produce a resultant are called the components of the resultant. a. either true or false c. neither true nor false b. false d. true PRIMING Components of a Vector In a two-dimensional coordinate system, any vector can be broken into x -component and y component. V = Vx , Vy For example, in the figure shown below, the vector v⃗ v→ is broken into two components, Vx and Vy . Let the angle between the vector and its x -component be θ. 61 Specialized Subject - STEM Downloaded from https://www.varsitytutors.com/hotmath/hotmath_help/topics/components-of-avector The vector and its components form a right angled triangle as shown below. Downloaded from https://www.varsitytutors.com/hotmath/hotmath_help/topics/components-ofa-vector In the above figure, the components can be quickly read. The vector in the component form is v⃗ =⟨4,5⟩v→=⟨4,5⟩ . The trigonometric ratios give the relation between magnitude of the vector and the components of the vector. cos = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑉𝑥 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑉 sin = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑉𝑦 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑉 62 Specialized Subject - STEM Vx = Vcos Vy = Vsin Using the Pythagorean Theorem in the right triangle with lengths vx and vy : 𝑉 = √𝑉𝑥 2 + 𝑉𝑦 2 PROCESSING Determining the Resultant and Direction of Multiple Vectors A = 50 N 300 N of E B = 25 N 650 S of W C = 45 N. 200 S of E 1. Draw the vectors in the Cartesian plane. 2. Compute the x and y components of each vector. Note the sign of each component based on the location in the Cartesian plane. 3. Add all the x-components and y-components. 4. Calculate the resultant and direction using the formula below. 𝑹 = √𝒙𝟐 + 𝒚𝟐 𝒚 = 𝒙 Vectors x- component y- component A 50 N cos 300 = 43.3 N 50 N sin 300 = 25 N B -25 N cos 650 = 10.6 N 25 N sin 650 = 22.7 N C 45 N cos 200 = 42.3 N -45 N sin 200 = -15.4 N x = 75 N y = 32.3 N 63 Specialized Subject - STEM F = √75 𝑁 2 + 32.2 𝑁 2 F = 81.62 N = 32.3 75 = 23.30 N of E F = 81.62 N 23.30 N of E UNDERSTANDING: Post-test Activity Resultant Vector of Typhoon Yolanda Identify the velocity of Typhoon Yolanda as it enters and exit the Philippine Area of Responsibility. Calculate the resultant velocity. Vector Velocity (m/s) x-component y-component 425 km E SE 64 km/h W 241 km/h W 34 km/h W 250 km/h W 314 km/h W 378 km/h W 314 km/h W 298 km/h W 64 Specialized Subject - STEM x _____________ = y _____________ = = ____________ Vx = ____________ Vy = ___________________ V = ___________ Post Test Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 1. In a coordinate system, a vector is oriented at angle with respect to the x-axis. The x component of the vector equals the vector’s magnitude multiplied by which trigonometric function? a. tan c. cot b. cos d. sin 2. A particular hurricane travels 678 km, 34.60 north of west before turning into a tropical storm. Find the northern displacement of the typhoon and the western displacement of the typhoon. a. 558 km west, 385 km north b. 385 km west, 558 km north c. 585 km west, 358 km north d. 468 km west, 468 km north For numbers 3-4 3. Two forces act on an object. One force is 6.0 N horizontally towards west. The second force is 8.0 N vertically towards south. Find the magnitude and direction of the resultant. a. 10N 53⁰ N of E c. 10N 53⁰ E of N b. 10N 53⁰ S of W d. 10N 53⁰ W of S 4. If the object is in equilibrium, find the magnitude and direction of the force that produces equilibrium. a. 10N, 53⁰ W of S c. 10N, 53⁰ E of N b. 10N, 53⁰ N of E d. 10N, 53⁰ S of W 65 Specialized Subject - STEM 5. Four members of the Main Street Bicycle Club meet at a certain intersection on Main Street. The members then start from the same location but travel in different directions. A short time later, displacement vectors for the four members are: A = 2 km W B = 1.6 km N C = 2.0 km E D = 2.4 km S What is the resultant displacement R of the members of the bicycle club: R = A + B + C + D? a. 0.8 km S b. 0.4 km 450 SE c. 3.6 km 370 NW d. 4 km S 6. Given the following components for vectors A–C, find the x- and y- components for the resultant R. a. +11, +11 b. +7, +7 c. –7, –7 d. +7, –11 7. Given the following components for vectors A–C, find the magnitude and direction for the resultant vector R. a. b. c. d. 8, 320° in standard position 10, 40° in standard position 8, 330° in standard position 10, 30° in standard position 66 Specialized Subject - STEM 8. Find the x- and y-components for a displacement vector that is 23.8 km and 45.0° south of east. a. b. c. d. - 12.61 km, +20.23 km +12.61 km, -20.23 km +16.8 km, –16.8 km –16.8 km, +16.8 km 9. A particular hurricane traveled 678 mi at 34.6° north of west before turning into a tropical storm. Find the northern displacement of the hurricane and the western displacement of the hurricane. a. 558 mi east, 385 mi north b. 385 mi west, 558 mi north c. 671 mi west, 27.12 mi north d. 468 mi west, 468 mi north 10. Find the x- and y-components to a vector that is 89.5 mm at 305° in standard position. a. –73.3 mm, 51.3 mm b. 51.3 mm, 73.3 mm c. 73.3 mm, 51.3 mm d. 85.95mm, 23.45mm 11. When resolving vectors into components or finding results __________ is/are more accurate than __________. a. geometric vector addition, geometric vector subtraction b. geometric techniques, mathematical techniques c. mathematical techniques, geometric techniques d. mathematical vector addition, mathematical vector subtraction 12. Resolve vector L into components Lx and Ly if the length of vector L is 15m and its reference angle is 200. a. 13.65 m, 6.15 m c. 14.1, 5.13 m b. 14 m, 5 m d. 14.2, 5.20 m 13. Which is not true about vector magnitude? a. it cannot be greater than the sums of magnitude of its component vectors. b. it cannot be negative c. it is scalar quantity d. trigonometry is necessary to compute it from component vectors 67 Specialized Subject - STEM 14. The vector resultant of an object’s change in position is the same at its displacement. a. either true or false c. neither true nor false b. false d. true 15. Two vectors that are added together to produce a resultant are called the components of the resultant. a. either true or false c. neither true nor false b. false d. true 68 Specialized Subject - STEM MODULE3Q1 KINEMATICS : MOTION ALONG A STRAIGHT LINE Lesson 2: Uniformly Accelerated Motion Overview: This module was designed and written with you in mind. It is here to help you master the Kinematics: Motion Along a Straight Line. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Convert a verbal description of a physical situation involving uniform acceleration in one dimension into a mathematical description Lesson Learning Objectives: After going through this module, you are expected to: ✓ convert a verbal description of a physical situation involving uniform acceleration in one dimension into a mathematical description PRE-TEST Direction: Read the questions properly, choose the correct answer and write it on the blanks provided before each item. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 1. Which of the following is acceleration? a. 10 m/s b. 46 km/h c. 50 cm2/s2 d.64 km/h/min 2. Acceleration is negative if speed is a. constant c. increasing b. decreasing d. neither increasing nor decreasing 69 Specialized Subject - STEM 3. The distance in meters traveled by a particle is related to time (t) in seconds by the equation of motion -S = 10 t +4 t2. What is the initial velocity of the body? a. 4 m/s b. 6 m/s c. 10 m/s d. 10 m/s2 4. The particle moves on the x-axis. When its acceleration is positive and increasing: a. its velocity must be positive b. its velocity must be negative c. it must be slowing down d. it must be speeding up e. none of the above must be true 5. Which of the following is correct when the distance of an object covered is directly proportional to time? a. constant acceleration c. uniform acceleration b. constant speed d. zero velocity 6. An object moving in the +x direction experiences an acceleration of +2.0 m/s2. This means the object a. travels 2.0 m in every second. b. is traveling at 2.0 m/s. c. is decreasing its velocity by 2.0 m/s every second. d. is increasing its velocity by 2.0 m/s every second. 7. A racing car accelerates uniformly from rest along a straight track. This track has markers spaced at equal distances along it from the start, as shown in the figure. The car reaches a speed of 140 km/h as it passes marker 2. Where on the track was the car when it was traveling at half this speed, that is at 70 km/h? 70 Specialized Subject - STEM a. before marker 1 b. At marker 1 c. Between marker 1 and marker 2 d. at marker 2 8. From the equations of distance, the correct one is a. Vf = Vi + 2as c. Vf2 = Vi2 + 2a b. Vf2 = Vi2 + as d. Vf2 = Vi2 + 2as 9. Correct equation of distance is a. Vi = Vf + at c. Vf = Vi + t b. Vf = Vi + at d. Vf = Vi + a 10. Jackson travels 2 km north, then 3 km east, and finally 2 km south. Which statement is true? a. Jackson’s displacement is 2 km west from his origin. b. Jackson is now 3 km east from where he started. c. Jackson’s displacement is 7 km. d. None of the above. 71 Specialized Subject - STEM PRIMING Downloaded from https://www.thoughtco.com/one-dimensional-kinematics-motion-straightline-2698879 Velocity in One-Dimensional Kinematics Velocity represents the rate of change of displacement over a given amount of time. The displacement in one-dimension is generally represented in regards to a starting point of x1 and x2. The time that the object in question is at each point is denoted as t1 and t2 (always assuming that t2 is later than t1, since time only proceeds one way). The change in a quantity from one point to another is generally indicated with the Greek letter delta, Δ, in the form of: Using these notations, it is possible to determine the average velocity (vav) in the following manner: vav = (x2 - x1) / (t2 - t1) = Δx / Δt If you apply a limit as Δt approaches 0, you obtain an instantaneous velocity at a specific point in the path. Such a limit in calculus is the derivative of x with respect to t, or dx/dt. Acceleration in One-Dimensional Kinematics 72 Specialized Subject - STEM Acceleration represents the rate of change in velocity over time. Using the terminology introduced earlier, we see that the average acceleration (aav) is: aav = (v2 - v1) / (t2 - t1) = Δx / Δt Again, we can apply a limit as Δt approaches 0 to obtain an instantaneous acceleration at a specific point in the path. The calculus representation is the derivative of v with respect to t, or dv/dt. Similarly, since v is the derivative of x, the instantaneous acceleration is the second derivative of x with respect to t, or d2x/dt2. Constant Acceleration In several cases, such as the Earth's gravitational field, the acceleration may be constant - in other words the velocity changes at the same rate throughout the motion. Using our earlier work, set the time at 0 and the end time as t (picture starting a stopwatch at 0 and ending it at the time of interest). The velocity at time 0 is v0 and at time t is v, yielding the following two equations: a = (v - v0)/(t - 0) v = v0 + at Applying the earlier equations for vav for x0 at time 0 and x at time t, and applying some manipulations (which I will not prove here), we get: x = x0 + v0t + 0.5at2 v2 = v02 + 2a (x - x0) x - x0 = (v0 + v) t/2 The above equations of motion with constant acceleration can be used to solve any kinematic problem involving motion of a particle in a straight line with constant acceleration. PROCESSING Motion with constant acceleration 73 Specialized Subject - STEM When an object moves with constant acceleration, the velocity increases or decreases at the same rate throughout the motion. The average acceleration equals the instantaneous acceleration when the acceleration is constant. A negative acceleration can indicate either of two conditions: Case 1: The object has a decreasing velocity in the positive direction. Case 2: The object has an increasing velocity in the negative direction. For example, a ball tossed up will be under the influence of a negative (downward) acceleration due to gravity. Its velocity will decrease while it travels upward (case 1); then, after reaching its highest point, the velocity will increase downward as the object returns to earth (case 2). Using v o (velocity at the beginning of time elapsed), v f (velocity at the end of the time elapsed), and t for time, the constant acceleration is Substituting the average velocity as the arithmetic average of the original and final velocities v avg = ( v o + v f )/2 into the relationship between distance and average velocity d = ( v avg)( t) yields. Substitute v f from Equation 1 into Equation 2 to obtain Finally, substitute the value of t from Equation 1 into Equation 2 for These four equations relate v o , v f , t, a, and d. Note that each equation has a different set of four of these five quantities. Table summarizes the equations for motion in a straight line under constant acceleration. 74 Specialized Subject - STEM Downloaded from https://www.cliffsnotes.com/study-guides/physics/classical mechanics/kinematics-in-one-dimension A special case of constant acceleration occurs for an object under the influence of gravity. If an object is thrown vertically upward or dropped, the acceleration due to gravity of −9.8 m/s 2 is substituted in the above equations to find the relationships among velocity, distance, and time. UNDERSTANDING: Post-test Activity 1 Three pairs of initial and final positions along an x-axis represent the location of objects at two successive times: 1. -3 m, +5 m 2. -3 m, -7 m 3. 7 m, -3 m a. Which pairs give a negative acceleration? b. Calculate the value of displacement in each case using vector notation. 75 Specialized Subject - STEM Post Test Multiple Choice. Choose the letter of the best answer. 1. Which of the following is acceleration? a. 10 m/s b. 46 km/h c. 50 cm2/s2 2. Acceleration is negative if speed is a. constant b. decreasing d.64 km/h/min c. increasing d. neither increasing nor decreasing 3. The distance in meters traveled by a particle is related to time (t) in seconds by the equation of motion -S = 10 t +4 t2. What is the initial velocity of the body? a. 4 m/s b. 6 m/s c. 10 m/s d. 10 m/s2 4. a. b. c. d. e. The particle moves on the x-axis. When its acceleration is positive and increasing: its velocity must be positive its velocity must be negative it must be slowing down it must be speeding up none of the above must be true 5. Which of the following is correct when the distance of an object covered is directly proportional to time? a. constant acceleration c. uniform acceleration b. constant speed d. zero velocity 6. An object moving in the +x direction experiences an acceleration of +2.0 m/s 2. This means the object a. travels 2.0 m in every second. b. is traveling at 2.0 m/s. c. is decreasing its velocity by 2.0 m/s every second. d. is increasing its velocity by 2.0 m/s every second. 76 Specialized Subject - STEM 7. A racing car accelerates uniformly from rest along a straight track. This track has markers spaced at equal distances along it from the start, as shown in the figure. The car reaches a speed of 140 km/h as it passes marker 2. Where on the track was the car when it was traveling at half this speed, that is at 70 km/h? a. before marker 1 b. At marker 1 c. Between marker 1 and marker 2 d. at marker 2 8. From the equations of distance, the correct one is a. Vf = Vi + 2as c. Vf2 = Vi2 + 2a b. Vf2 = Vi2 + as d. Vf2 = Vi2 + 2as 9. Correct equation of distance is a. Vi = Vf + at c. Vf = Vi + t b. Vf = Vi + at d. Vf = Vi + a 10. Jackson travels 2 km north, then 3 km east, and finally 2 km south. Which statement is true? a. Jackson’s displacement is 2 km west from his origin. b. Jackson is now 3 km east from where he started. c. Jackson’s displacement is 7 km. d. None of the above. 11. You drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. Your average speed over the 12 km drive will be a. greater than 70 km/h. b. equal to 70 km/h. 77 Specialized Subject - STEM c. less than 70 km/h. d. exactly 38 km/h. 12. Which of the following situations is impossible? a. An object has velocity directed east and acceleration directed west. b. An object has velocity directed east and acceleration directed east. c. An object has zero velocity but non-zero acceleration. d. An object has constant non-zero acceleration and changing velocity. 13. If the acceleration of an object is zero, then that object cannot be moving. a. either true or false b. False c. neither true nor false d. true 14. If the velocity of an object is zero, then that object cannot be accelerating. a. either true or false b. False c. neither true nor false d. true 15. An object moving in the +x direction experiences an acceleration of +5.0 m/s 2. This means the object a. travels 5.0 m in every second. b. is traveling at 5.0 m/s. c. is decreasing its velocity by 5.0 m/s every second. d. is increasing its velocity by 5.0 m/s every second. 78 Specialized Subject - STEM MODULE3Q1 KINEMATICS : MOTION ALONG A STRAIGHT LINE Lesson 2: Uniformly Accelerated Mot Overview: This module was designed and written with you in mind. It is here to help you master the Uniformly Accelerated Motion. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Solve for unknown quantities in equations involving one-dimensional uniformly accelerated motion , including free fall motion Lesson Learning Objectives: After going through this module, you are expected to: ✓ identify the given and unknown quantities of the given problem; ✓ derive the equation to be used to solve the given problem; solve for unknown quantities in equations involving one-dimensional uniformly accelerated motion. PRE-TEST Direction: Read the questions properly, encircle the correct answer. 1. An object moves with a constant acceleration of 5 m/s2. Which of the following statements is true? a. The object’s velocity stays the same b. The object moves 5 m each second c. The object’s acceleration increases by 5 m/s2 each second d. the object’s velocity increases by 5 m/s each second 2. A toy car moves 8 m in 4 s at the constant velocity. What is the car’s velocity? a. 1 m/s c. 3 m/s b. 2 m/s d. 4 m/s 79 Specialized Subject - STEM 3. If a total distance of 750 m is covered in a time interval of 25 s, the average speed is______. a. 3974 mph c. 30 mph b. 3.0 mph d. 30 m/s 4. If a person walked at 2 m/s for 12 s, he/she would travel a distance of______. a. 24 m c. 4 m b. 6 m d. None of the answers 5. How long would it take to travel 50 km travelling at a speed of 10 km/hr? a. 180 min c. 300 min b. 60 min d. 250 min 6. A car starts from point A, goes 50 km in a straight line to point B, immediately turns around, and returns to A. the time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is_____. a. 0 km/hr b. 50 km/hr c. 100 km/hr d. 200 km/hr 7. Still referring to the situation described in the previous question, what is the average speed of the car? a. 0 km/hr b. 50 km/hr c. 100 km/hr d. 200 km/hr 8. A train moves at a constant velocity of 50 km/h. How far will it move in 0.5 h? a. 10 km c. 25 km b. 20 km d. 45 km 80 Specialized Subject - STEM 9. A boat can move at a constant velocity of 8 km/h in still water. How long will it take for the boat to move 24 km? a. 2 h c. 4 h b. 3 h d. 6 h 10. A bicyclist moves at a constant speed of 4 m/s. How long it will take for the bicyclist to move 36 m? a. 3 s c. 9 s b. 6 s d. 12 s PRIMING Uniformly Accelerated Motion (UAM) is motion of an object where the acceleration is constant. In other words, the acceleration remains uniform; the acceleration is equal to a number and that number does not change as a function of time. Examples of objects in UAM: • A ball rolling down an incline. • A person falling from a plane. • A bicycle on which you have applied the brakes. • A ball dropped from the top of a ladder. • A toy baby bottle released from the bottom of a bathtub. (Technically, because of friction and a non-constant gravitational field, etc., they are not quite Uniformly Accelerated Motion, however, at this point we will treat them as if they are, because it is close enough, for now.) These are the equations that describe an object in Uniformly Accelerated Motion: Vf = Vi + a t x = Vi t + ½ a t2 Vf2 = Vi2 + 2a x 81 Specialized Subject - STEM ∆𝒙 = 𝟏 (𝑽𝒇 + 𝑽𝒊)∆𝒕 𝟐 There are 5 variables in the UAM equations: Vf – final velocity Vi – initial velocity a – acceleration x – displacement t – change in time PROCESSING Uniformly Accelerated Motion Super Problem A ball is thrown upward at 25 m/s from the ground. 1. What is the initial velocity of the ball? 2. What is the acceleration of the ball? 3. What is the ball’s velocity after 2 seconds? 4. What is the ball’s velocity after 4 seconds? 5. What is the maximum height of the ball? 6. How long until the ball hits the ground? 7. When is the magnitude of the velocity 5 m/s? 8. What distance has the ball travelled after 5 seconds? 9. What is the average velocity and average speed of the ball after 5 seconds? 10. Another ball is thrown one second later. What speed does it need to hit the ground simultaneously with the first ball? Solution: 1. Viy = +25 m/s 2. ay = - 9.8 m/s2 3. The final velocity is Vfy – Viy = ay t Vfy = Viy + ay t 82 Specialized Subject - STEM = 25 m/s + (-9.9 m/s2) (2s) = 5.4 m/s 4. The final velocity is Vfy – Viy = ay t Vfy = Viy + ay t = 25 m/s + (-9.9 m/s2) (4s) = -14.2 m/s 5. At the highest point, what is the velocity? Vfy2 – Viy2 = 2 ay y 0 – Viy2 = 2 ay y y = -Viy2 /2ay = -25 m/s2 /2 (-9.8 m/s2) = 31.9 m 6. What is the ball’s displacement when it returns to the ground? y = Viy t + ½ ay (t)2 0 = Viy t + ½ ay (t)2 = t (Viy + ½ ay t) t = −2 𝑉𝑖𝑦 𝑎𝑦 𝑚 = −2 (25 𝑠 ) 𝑚 − 9.8𝑠2 = 5.10 𝑠 How long does it take for the ball to reach its maximum height? 7. The velocity has magnitude 5 m/s when its value is +5 m/s and −5 m/s. For +5 m/s, Vfy – Viy = ay t t = Vfy - Viy / ay = 5 m/s - 25 m/s / 9.8 m/s2 = 2.04 s For -5 m/s t = Vfy - Viy / ay = -5 m/s - 25 m/s / 9.8 m/s2 = 3.06 s 83 Specialized Subject - STEM Notice these times are equidistant from the time it takes to reach the highest point, 2.55 s. The velocities are symmetric about the highest point. 8. The ball is descending at 5 s. The position at 5 seconds is y = Viy t + ½ ay (t)2 = (25 m/s) (5s) + ½ (-9.8 m/s2) = 2.5 m The ball rises to a maximum height of 31.9 m and falls to a height of 2.5 m, a distance of (31.9 m – 2.5 m) = 29.4 m below the highest point. The total distance travelled is d = 31.9 m + 29.4 m = 61.3 m 9. The average velocity is the displacement divided by the elapsed time Vave y = y /t = 2.5 m / 5s = 0.5 m/s The average speed of the distance divided by time Speed = d /t = 61.3 m /5s = 12.3 m/s 10. The second ball must be in the air 1 second shorter than the first ball. t2 = t1 – 1s = 5.10 s – 1.00 s = 4.10 s Again when it hits the ground, y = 0 y = Viy t + ½ ay (t)2 0 = Viy t + ½ ay (t)2 = t (Viy + ½ ay t) Viy = -1/2 ay t = -1/2 (-9.8 m/s2) (4.10 s) = 20.1 m/s 84 Specialized Subject - STEM UNDERSTANDING: Post-test Problem Solving: Identify the given quantities. Identify the known quantities. Derive the equation to solve the given problem. 1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifted off the ground. Determine the distance traveled before takeoff. 2. A car starts from rest and accelerated uniformly over a time of 5.21 seconds for distance of 110 m. Determine the acceleration of the car. 3. If an object falls for 2.60 seconds in a building, what will be its final velocity and how far will it fall? 4. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 s. Determine the acceleration of the car and the distance traveled. 5. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike. 6. An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this place will be 65 m/s. Assuming the minimum acceleration, what is the minimum allowed length for the runway? 7. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration). 8. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840. Determine the acceleration of the bullet (assume a uniform acceleration) 9. The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below. 10. A body moving along a straight line with a velocity of 40 m/s undergoes an acceleration of 4 m/s2. After 10 s its speed will be. 85 Specialized Subject - STEM Post Test Multiple Choice. Choose the letter of the best answer. 1. An object moves with a constant acceleration of 5 m/s2. Which of the following statements is true? a. The object’s velocity stays the same b. The object moves 5 m each second c. The object’s acceleration increases by 5 m/s2 each second d. the object’s velocity increases by 5 m/s each second 2. A toy car moves 8 m in 4 s at the constant velocity. What is the car’s velocity? a. 1 m/s c. 3 m/s b. 2 m/s d. 4 m/s 3. If a total distance of 750 m is covered in a time interval of 25 s, the average speed is______. a. 3974 mph c. 30 mph b. 3.0 mph d. 30 m/s 4. If a person walked at 2 m/s for 12 s, he/she would travel a distance of______. a. 24 m c. 4 m b. 6 m d. None of the answers 5. How long would it take to travel 50 km travelling at a speed of 10 km/hr? a. 180 min c. 300 min b. 60 min d. 250 min 6. A car starts from point A, goes 50 km in a straight line to point B, immediately turns around, and returns to A. the time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is_____. a. 0 km/hr c. 100 km/hr b. 50 km/hr d. 200 km/hr 7. Still referring to the situation described in the previous question, what is the average speed of the car? a. 0 km/hr c. 100 km/hr b. 50 km/hr d. 200 km/hr 8. A train moves at a constant velocity of 50 km/h. How far will it move in 0.5 h? a. 10 km c. 25 km b. 20 km d. 45 km 86 Specialized Subject - STEM 9. A boat can move at a constant velocity of 8 km/h in still water. How long will it take for the boat to move 24 km? a. 2 h c. 4 h b. 3 h d. 6 h 10. A bicyclist moves at a constant speed of 4 m/s. How long it will take for the bicyclist to move 36 m? a. 3 s c. 9 s b. 6 s d. 12 s 11. A bicyclist covers 60 miles between 2 pm and 6 pm. What was his average speed? a. 15 mph c. 45 mph b. 30 mph d. 60 mph 12. A Ronda takes ten minutes to go from milepost 71 to milepost 81. A Toyota takes fifteen minutes to go from milepost 65 to milepost 80. Which car has the higher average speed? a. the Ronda. b. the Toyota. c. The average speeds are the same. d. We need to know the accelerations to answer the question. e. Not enough information is given to be able to say. 13. What average speed, most nearly, is required to run a mile (1.6 km), in 4 minutes? a. 4.0 m/s c. 40.0 m/s b. 7.0 m/s d. 70 m/s 14. If a car requires 30 seconds to accelerate from zero to 90 km per hour, its average acceleration is, most nearly, a. .8 m/s2 c. 80 m/s2 b. 8 m/s2 d. 800 m/s2 15. A car initially traveling north at 5 m/s has a constant acceleration of 2 m/s2 northward. How far does the car travel in the first 10 s? a. 20m c. 100 m b. 50m d. 150 m 87 Specialized Subject - STEM MODULE3Q1 KINEMATICS : MOTION ALONG A STRAIGHT LINE Lesson 2: Uniformly Accelerated Motion Overview: This module was designed and written with you in mind. It is here to help you master the Motion Along a Straight Line. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Interpret displacement and velocity, respectively, as areas under velocity vs. time and acceleration vs. time curves, Interpret velocity and acceleration, respectively, as slopes of position vs. time and velocity vs. time curves, Construct velocity vs. time and acceleration vs. time graphs, respectively, corresponding to a given position vs. time-graph and velocity vs. time graph and vice versa Lesson Learning Objectives: After going through this module, you are expected to: ✓ interpret displacement and velocity, respectively, as areas under velocity vs. time and acceleration vs. time curves; ✓ understand displacement, velocity and acceleration; ✓ understand motion graphs (v vs. t and a vs. t); and ✓ be able to interpret motion graphs and to make prediction PRE-TEST Direction: Read the questions properly, encircle the correct answer. For numbers 1-2 The following graph represents the position as a function of time for a moving object. Use this graph to answer questions 1-2. 88 Specialized Subject - STEM 1. Which of the following is TRUE? a. The object increases its velocity b. The object decreases its velocity c. The object’s velocity stays unchanged d. The object stays at rest 2. What is the velocity of the object? a. 4 m/s c. 5 m/s b. 8 m/s d. 40 m/s For numbers 3-4 The following graph represents the position as a function of time of a moving object. Use this graph to answer questions 6 and 7. 3. What is the initial position of the object? a. 2 m b. 4 m c. 6 m d. 8 m 89 Specialized Subject - STEM 4. What is the velocity of the object? a. 2 m/s b. 4 m/s c. 6 m/s d. 8 m For Numbers 5-6. The following graph represents the position as a function of time of a moving object. Use this graph for questions 5 and 6. 5. What is the initial position of the object? a. 4 m b. 6 m c. 8 m d. 10 m 6. What is the velocity of the object? a. 5 m/s b. -5 m/s c. 10 m/s d. -10 m/s Time, t (s) Velocity, v (m/s) 0 0.0 1 2.0 2 4.0 3 6.0 4 8.0 5 10.0 90 Specialized Subject - STEM 6 12.0 7 14.0 7. The velocity –time graphs of the motion is a. horizontal line ` b. parabola and straight line. c straight lines of different shapes d. straight line and hyperbola 8. The slope of the graph between t=0.0 s and t= 6.0 s is a. 2 m/s2 b. 4 m/s2 c. 6m/s2 d. 8 m/s2 9. The slope of the line between t= 7.0 s and t = 10.0 s indicates a. decelerated motion b. uniformly accelerated motion c. uniform speed d. uniform velocity 10. The distance that the car has traveled during the first 6.0 s is a. 36 m b. 46 m c. 54 m d. 72 m PRIMING 91 Specialized Subject - STEM What does the area represent on velocity graph? The area under a velocity graph represents the displacement of the object. To see why, consider the following graph of motion that shows an object maintaining a constant velocity of 6 meters per second for a time of 5 seconds. V (m/s) 7 6 5 4 3 2 1 t (s) 1 2 3 4 5 6 7 To find the displacement during this time interval, we could use this formula 𝒎 ∆𝒙 = 𝒗 ∆𝒕 = (𝟔 𝒔 ) (𝟓 𝒔) = 𝟑𝟎 𝒎 which gives a displacement of 30 m. Now we're going to show that this was equivalent to finding the area under the curve. Consider the rectangle of area made by the graph as seen above. 92 Specialized Subject - STEM V (m/s) 7 6 5 4 3 2 1 t (s) 1 2 3 4 5 Downloaded from https://www.khanacademy.org/science/ap-physics-1/ap-one-dimensionalmotion/average-and-instantaneous-acceleration/a/what-are-velocity-vs-time-graphs The area of this rectangle can be found by multiplying height of the rectangle, 6 m/s, times its width, 5 s, which would give area = height x width = 6 m/s x 5 s = 30 m This is the same answer we got before for the displacement. The area under a velocity curve, regardless of the shape, will equal the displacement during that time interval. area under curve = displacement What does the area represent on an acceleration graph? The area under an acceleration graph represents the change in velocity. In other words, the area under the acceleration graph for a certain time interval is equal to the change in velocity during that time interval. area = V 93 Specialized Subject - STEM It might be easiest to see why this is the case by considering the example graph below which shows a constant acceleration of 4 m/s2 for a time of 9s. a (m/s2) 4 3 2 1 t (s) 1 2 3 4 5 6 7 8 9 Downloaded from https://www.khanacademy.org/science/ap-physics-1/ap-one-dimensionalmotion/average-and-instantaneous-acceleration/a/what-are-velocity-vs-time-graphs If we multiply both sides of the definition of acceleration, a = ∆𝑉 ∆𝑡 by the change in time t, we get V = a t. Plugging in the acceleration 4 m/s2 and the interval 9s we can find the change in velocity: V = a t = (4 m/s2) (9 s) = 36 m/s 94 Specialized Subject - STEM a (m/s2) 4 3 2 1 t (s) 1 2 3 4 5 6 7 8 9 Downloaded from https://www.khanacademy.org/science/ap-physics-1/ap-one-dimensionalmotion/average-and-instantaneous-acceleration/a/what-are-velocity-vs-time-graphs The area can be found by multiplying height times width. The height of this rectangle is a = 4 m/s2, and the width is 9s. So, finding the area also gives you the change in velocity. a = 4 m/s2 x 9 s = 36 m/s The area under any acceleration graph for a certain time interval gives the change in velocity for that time interval. PROCESSING Finding the displacement of the go-kart between t=0 and t=7s. 95 Specialized Subject - STEM We can find the displacement of the go-kart by finding the area under the velocity graph. The graph can be thought of as being a rectangle (between t = 0s and t = 3s) and a triangle (between t = 3s and t = 7s). Once we find the area of these shapes and add them, we will get the total displacement. V (m/s) 7 6 5 4 3 2 1 t (s) 1 2 3 4 5 6 7 Downloaded from https://www.khanacademy.org/science/ap-physics-1/ap-one-dimensionalmotion/average-and-instantaneous-acceleration/a/what-are-velocity-vs-time-graphs The area of the rectangle is found by area = h x w = 6 m/s x 3 s = 18 m The area of the triangle is found by 𝒂𝒓𝒆𝒂 = 𝟏 𝟐 𝟏 bh = 𝟐 (4 s) (6 m/s) = 12 m Adding these two areas together give the total displacement. total area = 18 m + 12 m = 30 m total displacement = 30m Race car acceleration A confident race car driver is cruising at a constant velocity of 20 m/s. As she nears the finish line, the race car driver starts to accelerate. The graph shown below gives the acceleration of the race car as it starts to speed up. Assume the race car had a velocity of 20 m/s at time t = 0 s. 96 Specialized Subject - STEM What is the velocity of the race car after the 8 seconds of acceleration shown in the graph? a (m/s2) 6 5 4 3 2 1 t (s) 1 2 3 4 5 6 7 8 Downloaded from https://www.khanacademy.org/science/ap-physics-1/ap-one-dimensionalmotion/average-and-instantaneous-acceleration/a/what-are-velocity-vs-time-graphs We can find the change in velocity by finding the area under the acceleration graph. v = area = ½ bh = ½ (8s) (6 m/s2) = 24 m/s But this is just the change in velocity during the time interval. We need to find the final velocity. We can use the definition of the change in velocity, V = Vf - Vi V = 24 m/s Vf – Vi = 24 m/s Vf – 20 m/s = 24 m/s Vf = 24 m/s + 20 m/s Vf = 44 m/s 97 Specialized Subject - STEM The final velocity of the race car was 44 m/s. UNDERSTANDING: Post-test Activity 1 1. A sailboat is sailing in a straight line with a velocity of 10 m/s. Then at time t=0s, a stiff wind blows causing the sailboat to accelerate as seen in the diagram below. What is the velocity of the sailboat after the wind has blown for 9 seconds? a (m/s2) 4 3 2 1 t (s) 1 2 3 4 5 6 7 8 9 Downloaded from https://www.khanacademy.org/science/ap-physics-1/ap-one-dimensionalmotion/average-and-instantaneous-acceleration/a/what-are-velocity-vs-time-graphs 2. A windsurfer is traveling along a straight line, and her motion is given by the velocity graph below. 98 Specialized Subject - STEM Select all of the following statements that are true about the speed and acceleration of the windsurfer. (A) Speed is increasing. (B) Acceleration is increasing. (C) Speed is decreasing. (D) Acceleration is decreasing. V (m/s) 7 6 5 4 3 2 1 t (s) 1 2 3 4 5 6 7 8 9 Downloaded from https://www.khanacademy.org/science/ap-physics-1/ap-one-dimensionalmotion/average-and-instantaneous-acceleration/a/what-are-velocity-vs-time-graphs Post Test 99 Specialized Subject - STEM Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. For numbers 1-2 The following graph represents the position as a function of time for a moving object. Use this graph to answer questions 1-2. 1. Which of the following is TRUE? a. The object increases its velocity b. The object decreases its velocity c. The object’s velocity stays unchanged d. The object stays at rest 2. What is the velocity of the object? a. 4 m/s b. 8 m/s c. 5 m/s d. 40 m/s For numbers 3-4 The following graph represents the position as a function of time of a moving object. Use this graph to answer questions 3 and 4. 100 Specialized Subject - STEM 3. What is the initial position of the object? a. 2 m b. 4 m c. 6 m d. 8 m 4. What is the velocity of the object? a. 2 m b. 4 m c. 6 m d. 8 m For Numbers 5-6. The following graph represents the position as a function of time of a moving object. Use this graph for questions 5 and 6. 5. What is the initial position of the object? a. 4 m b. 6 m c. 8 m d. 10 m 101 Specialized Subject - STEM 6. What is the velocity of the object? a. 5 m/s b. -5 m/s c. 10 m/s d. -10 m/s Time, t (s) Velocity, v (m/s) 0 0.0 1 2.0 2 4.0 3 6.0 4 8.0 5 10.0 6 12.0 7 14.0 7. The velocity –time graphs of the motion is ` a. horizontal line b. parabola and straight line. c straight lines of different shapes d. straight line and hyperbola 8. The slope of the graph between t=0.0 s and t= 6.0 s is a. 2 m/s b. 4 m/s c. 6m/s d. 8 m/s 9. The slope of the line between t= 7.0 s and t = 10.0 s indicates a. b. decelerated motion uniformly accelerated motion 102 Specialized Subject - STEM c. uniform speed d. uniform velocity 10. The distance that the car has traveled during the first 6.0 s is a. 36 m b. 46 m c. 54 m d. 72 m Nos. 11-12 Construct the velocity vs. time graph based on the given data table. For numbers 13-14 The graph represents the relationship between velocity and time for an object moving in a straight line. Use this graph to answer questions 13 and 14. 103 Specialized Subject - STEM 13. Which of the following statements is true? a. b. c. d. The object speeds up The object slows down The object moves with a constant velocity The object stays at rest 14. What is the velocity of the object at 5 s? a. 0 m/s c. 5 m/s b. 3 m/s d. 4 m/s 15. The graph represents the relationship between velocity and time for an object moving in a straight line. What is the traveled distance of the object at 9 s? a. 10 m c. 36 m b. 24 m d. 48 m 104 Specialized Subject - STEM MODULE3Q1 KINEMATICS : MOTION ALONG A STRAIGHT LINE Lesson 2: Uniformly Accelerated Mot Overview: This module was designed and written with you in mind. It is here to help you master the Uniformly Accelerated Motion. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Describe motion using the concept of relative velocities in 1D and 2D Deduce the consequences of the independence of vertical and horizontal components of projectile motion Calculate range, time of flight, and maximum heights of projectiles Lesson Learning Objectives: After going through this module, you are expected to: ✓ identify the given and unknown quantities of the given problem; ✓ derive the equation to be used to solve the given free-fall problems; ✓ use the fact that the magnitude of acceleration due to gravity on the Earth’s surface is nearly constant and approximately 9.8 m/s2 in free- fall problems; and derive the equation to be used to solve the given 1D uniform Acceleration problems. PRE-TEST Direction: Read the questions properly, encircle the correct answer. 1. When we say that light objects and heavy objects fall at the same rate, what assumption(s) are we making? a. They have the same shape. b. They have the same size. c. They have surfaces with similar air resistances. d. They are falling in a vacuum. 105 Specialized Subject - STEM 2. When an object was thrown upwards reaches its highest point, which is TRUE? a. The acceleration switches from positive to negative. b. The acceleration is zero. c. The total displacement is zero. d. The velocity is zero. 3. An object is allowed to fall freely near the surface of a planet. The object falls 54 meters in the first 3 seconds after it is released. The acceleration due to gravity of that planet is______. a. 6 m/s2 c. 27 m/s2 b. 12 m/s2 d. 108 m/s2 4. Pedro was angry and wishes to drop an egg onto the head of Juan. He stations himself in a building window 19.6 m above the level of Juan’s head. Determine how many seconds before Juan is directly beneath him that he will have to drop the egg in order to get the desired plat? a. 1.5 seconds c. 2.5 seconds b. 2.0 seconds d. 3.0 seconds For numbers 5-7 5. A stone is thrown from the top of the building with an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. Determine the time at which the stone reaches its maximum height. a. 2.0 s c. 4.0 s b. 3.0 s d. 5.0 s 6. Determine the maximum height of the stone above the rooftop. a. 1.6 m c. 15.9m b. 10.6 m d. 20.4 m 7. Determine the time at which the stone returns to the level of the thrower. a. 4.0 s c. 8.0 s 106 Specialized Subject - STEM b. 6.0 s d. 10.0 s 8. A ball is in free fall. Its acceleration is: a. downward during both ascent and descent b. downward during ascent and upward during descent c. upward during ascent and downward during descent d. upward during both ascent and descent 9. A ball is in free fall. Upward is taken to be in positive direction. The displacement of the ball during a short time interval is: a. positive during ascent and negative during descent. b. positive during both ascent and descent c. negative during ascent and positive during descent d. negative during both ascent and descent. 10. A freely falling body has a constant acceleration of 9.8 m/s2. This means that; a. the acceleration of the body increases by 9.8 m/s2 during each second b. the body falls 9.8 m during each second c. the body falls 9.8 during the first second only d. the speed of the body increases by 9.8 m/s during each second PRIMING Free Fall The motion of falling objects is the simplest and most common example of motion with changing velocity. If a coin and a piece of paper are simultaneously dropped side by side, the paper takes much longer to hit the ground. However, if you crumple the paper into a compact ball and drop the items again, it will look like both the coin and the paper hit the floor simultaneously. This is because the amount of force acting on an object is a function of not only its mass, but also area. Free fall is the motion of a body where its weight is the only force acting on an object. 107 Specialized Subject - STEM Galileo also observed this phenomenon and realized that it disagreed with the Aristotle principle that heavier items fall more quickly. Galileo then hypothesized that there is an upward force exerted by air in addition to the downward force of gravity. If air resistance and friction are negligible, then in a given location (because gravity changes with location), all objects fall toward the center of Earth with the same constant acceleration, independent of their mass, that constant acceleration is gravity. Air resistance opposes the motion of an object through the air, while friction opposes motion between objects and the medium through which they are traveling. The acceleration of free-falling objects is referred to as the acceleration due to gravity gg. As we said earlier, gravity varies depending on location and altitude on Earth (or any other planet), but the average acceleration due to gravity on Earth is 9.8 m/s2. This value is also often expressed as a negative acceleration in mathematical calculations due to the downward direction of gravity. Equations The best way to see the basic features of motion involving gravity is to start by considering straight up and down motion with no air resistance or friction. This means that if the object is dropped, we know the initial velocity is zero. Once the object is in motion, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration, gg. The kinematic equations for objects experiencing free fall are: V = V0 – gt y = y0 + V0 t – ½ gt2 V2 = V02 -2g (y-y0) Where V = velocity, g = gravity, t = time and y = vertical displacement. PROCESSING An hour and thirty-one minutes after launch, my pressure altimeter halts at 103,300 feet. At ground control the radar altimeters also have stopped on readings of 102,800 feet, the figure that we later agree upon as the more reliable. It is 7 o'clock in the morning, and I have reached float altitude. At zero count I step into space. No wind whistles or billows my clothing. I have absolutely no sensation of the increasing speed with which I fall. Though my stabilization chute opens at 96,000 feet, I accelerate for 6,000 feet more before hitting a peak of 614 miles an hour, nine-tenths the speed of sound at my altitude. An Air Force camera 108 Specialized Subject - STEM on the gondola took this photograph when the cotton clouds still lay 80,000 feet below. At 21,000 feet they rushed up so chillingly that I had to remind myself they were vapor and not solid. For most skydivers, the acceleration experienced while falling is not constant. As a skydiver's speed increases, so too does the aerodynamic drag until their speed levels out at a typical terminal velocity of 55 m/s (120 mph). Air resistance is not negligible in such circumstances. The story of Captain Kittinger is an exceptional one, however. At the float altitude where his dive began, the Earth's atmosphere has only 1.5% of its density at sea level. It is effectively a vacuum and offers no resistance to a person falling from rest. The acceleration due to gravity is often said to be constant, with a value of 9.8 m/s2. Over the entire surface of the Earth up to an altitude of 18 km, this is the value accurate to two significant digits. In actuality, this "constant" varies from 9.81 m/s2 at sea level to 9.75 m/s2 at 18 km. At the altitude of Captain Kittinger's dive, the acceleration due to gravity was closer to 9.72 m/s2. Given this data it is possible to calculate the maximum speed of Captain Kittinger during his descent. First we will need to convert the altitude measurements. To save calculation time we will only convert the change in altitude and not each altitude. Given that he stepped out of the gondola at 102,800 feet, fell freely until 96,000 feet, and then continued to accelerate for another 6,000 feet; the distance over which he accelerated uniformly was… 102,800 − 96,00 + 6,000 = 12,800 feet 12,800 feet 1609 m 1 5280 feet = 3900 m It's now just a matter of choosing the correct formula and plugging in the numbers. v0 = 0 m/s a= 9.72 m/s2 ∆s = 3900 m v= ? 2 2 v = v0 + 2a∆s v= √(2a∆s) v= √(2(9.72 m/s2)(3900 m)) v= 275 m/s This result is amazingly close to the value recorded in Kittinger's report. 614 mile 1609 m 1 hour = 274 m/s 109 Specialized Subject - STEM 1 hour 1 mile 3600 s As one would expect the actual value is slightly less than the theoretical value. This agrees with the notion of a small but still non-zero amount of drag. UNDERSTANDING: Post-test Identify the given quantities, unknown quantities and equations to solve the given problem. 1. (a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air? 2. A small rock is thrown vertically upward with a speed of 22.0 m/s from the edge of the roof of a 30.0-m-tall building. The rock doesn’t hit the building on its way back down and lands on the street below. Ignore air resistance. (a) What is the speed of the rock just before it hits the street? (b) How much time elapses from when the rock is thrown until it hits the street? 3. A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler’s hand? 4. You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/s. (a) What is the speed of the putty just before it strikes the ceiling? (b) How much time from when it leaves your hand does it take the putty to reach the ceiling? 5. A tennis ball on Mars, where the acceleration due to gravity is 0.379g and air resistance is negligible, is hit directly upward and returns to the same level 8.5 s later. (a) How high above its original point did the ball go? (b) How fast was it moving just after it was hit? (c) Sketch graphs of the ball’s vertical position, vertical velocity, and vertical acceleration as functions of time while it’s in the Martian air. Post Test Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 1. When we say that light objects and heavy objects fall at the same rate, what assumption(s) are we making? 110 Specialized Subject - STEM a. They have the same shape. b. They have the same size. c. They have surfaces with similar air resistances. d. They are falling in a vacuum. 2. When an object was thrown upwards reaches its highest point, which is TRUE? a. The acceleration switches from positive to negative. b. The acceleration is zero. c. The total displacement is zero. d. The velocity is zero. 3. An object is allowed to fall freely near the surface of a planet. The object falls 54 meters in the first 3 seconds after it is released. The acceleration due to gravity of that planet is______. a. 6 m/s2 c. 27 m/s2 b. 12 m/s2 d. 108 m/s2 4. Pedro was angry and wishes to drop an egg onto the head of Juan. He stations himself in a building window 19.6 m above the level of Juan’s head. Determine how many seconds before Juan is directly beneath him that he will have to drop the egg in order to get the desired plat? a. 1.5 seconds c. 2.5 seconds b. 2.0 seconds d. 3.0 seconds For numbers 5-7 5. A stone is thrown from the top of the building with an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down. Determine the time at which the stone reaches its maximum height. a. 2.0 s c. 4.0 s b. 3.0 s d. 5.0 s 111 Specialized Subject - STEM 6. Determine the maximum height of the stone above the rooftop. a. 1.6 m c. 15.9m b. 10.6 m d. 20.4 m 7. Determine the time at which the stone returns to the level of the thrower. a. 4.0 s c. 8.0 s b. 6.0 s d. 10.0 s 8. A ball is in free fall. Its acceleration is: a. downward during both ascent and descent b. downward during ascent and upward during descent c. upward during ascent and downward during descent d. upward during both ascent and descent 9. A ball is in free fall. Upward is taken to be in positive direction. The displacement of the ball during a short time interval is: a. positive during ascent and negative during descent. b. positive during both ascent and descent c. negative during ascent and positive during descent d. negative during both ascent and descent. 10. A freely falling body has a constant acceleration of 9.8 m/s 2. This means that; a. the acceleration of the body increases by 9.8 m/s2 during each second b. the body falls 9.8 m during each second c. the body falls 9.8 during the first second only d. the speed of the body increases by 9.8 m/s during each second 112 Specialized Subject - STEM 11. If the contraction of the left ventricle lasts 250 m/s and the speed of blood flow in the aorta (the large artery leaving the heart) is 0.80 m/s at the end of the contraction, what is the average acceleration of a red blood cell as it leaves the heart? a. 0.32 m/s2 b. 3.2 m/s2 c. 31 m/s2 d. 310 m/s2 12. If the aorta (diameter da) branches into to equal-sized arteries with a combined area equal to that of the aorta, what is the diameter of one of the branches? a. 1da b. da /12 c. 2da d da/2 13. Rocket-powered sleds are used to test the human response to acceleration. If a rocketpowered sled is accelerated to a speed of 444 m/s in 1.83 s, what is the acceleration and what is the distance that the sled travels? a. a = 200 m/s2, d = 400 m b. a = 243 m/s2, d = 406 m c. a = 250 m/s2, d = 410 m d. a = 260 m/s2, d = 420 m 14. A feather, initially at rest, is released in a vacuum 12 m above the surface of the earth. Which of the following statements is correct? a. The acceleration of the feather decreases until terminal velocity is reached. b. The acceleration of the feather remains constant during the fall c. The acceleration of the feather increases during the fall. d. The maximum velocity of the feather is 9.8 m/s2. 113 Specialized Subject - STEM 15. A lunar lander is descending toward the moon’s surface. Until the lander reaches the surface, its height above the surface of the moon is given by y1t2 = b - ct + dt2, where b = 800 m is the initial height of the lander above the surface, c = 60.0 m/s, and d = 1.05 m/s2. What is the initial velocity of the lander, at t = 0? a. 0 m/s b. 50 m/s c. 60 m/s d. 100 m/s 114 Specialized Subject - STEM MODULE3Q1 KINEMATICS : MOTION ALONG A STRAIGHT LINE Lesson 2: Quantities in Circular Motion Overview: This module was designed and written with you in mind. It is here to help you master the accuracy and precision. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Infer quantities associated with circular motion such as tangential velocity, centripetal acceleration, tangential acceleration, radius of curvature Solve problems involving two dimensional motion in contexts such as, but not limited to ledge jumping, movie stunts, basketball, safe locations during firework displays, and Ferris wheels Lesson Learning Objectives: After going through this module, you are expected to: ✓ Explain the differences of quantities such as tangential velocity, centripetal acceleration, tangential acceleration, and radius of curvature present in an object Pin circular motion ✓ Use equations in finding position, velocity, and acceleration of objects in circular motion ✓ Solve problems involving centripetal acceleration of an object moving in circular path PRE-TEST Direction: Read the questions properly, encircle the correct answer. 1. Which of the following is the motion along the circle and at any point is always tangent to the circle? a. Tangential acceleration b. Tangential velocity c. Centripetal acceleration d. Radius of curvature 2. Which of the following shows the concept in an object that is moving in a circular motion with the acceleration is always towards the center of the circle? a. Tangential acceleration b. Tangential velocity c. Centripetal acceleration d. Radius of curvature 115 Specialized Subject - STEM 3. Which of the following is the measure how the tangential velocity changes along the time? a. Tangential acceleration b. Centripetal acceleration c. Centripetal acceleration d. Radius of curvature 4. Which of the following is the distance from the vertex to the center of curvature? a. Tangential acceleration b. Centripetal acceleration c. Centripetal acceleration d. Radius of curvature 5. An object moving along circular motion with acceleration vector pointing towards the center of the circle is known as centripetal acceleration. a. True b. sometimes c. false d. maybe not PRIMING Quantities of Circular Motion 1. 2. 3. 4. Tangential Velocity Tangential Acceleration Centripetal Acceleration Radius of Curvature 116 Specialized Subject - STEM PROCESSING Knowing that a quantity with both direction and magnitude is called a vector. An example of vector is velocity, wherein velocity is an example of linear motion where the rate of change in objects position with respect to time. 𝑠𝑝𝑒𝑒𝑑 /𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 --------equation1 (speed along linear motion) While the direction of the circular motion is the circumference of a circle is two pie times radius. 𝑐𝑖𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 = 2𝜋 ∗ 𝑟𝑎𝑑𝑖𝑢𝑠-------equation2 117 Specialized Subject - STEM Substitute equation 2 and 1 𝑠𝑝𝑒𝑒𝑑 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 = 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 = 2𝜋𝑟 𝑡 ----equation 3 (speed along circular motion) TANGENTIAL VELOCITY Tangential velocity is the velocity measured of an object moving along the edge of a circle and direction is always along with the tangent line of the circle at any given point. 𝑉𝑇 = 2𝜋𝑟 ∆𝑠 = 𝑡 ∆𝑡 Where VT = tangential velocity r = radius t = time Fig 1. Velocity vector direction at any point in circular path is always tangent to the circle Because of this understanding, the tangential velocity is related to angular velocity where; Object’s position with respect to time The tangential velocity is velocity with respect to time and the radius ∆𝑠 𝑟 ∙ ∆𝜃 𝑉𝑇 = = = 𝜔∙𝑟 ∆𝑡 ∆𝑡 𝜔 – angular velocity 𝜃 𝜔= 𝑡 measured by the angular of the wheel S = r.𝜃 Where ∆𝑠 – change in position ∆𝑡 – change in time 𝑠=𝑟∙𝜃 𝜔 – angular velocity r - radius Example 118 Specialized Subject - STEM Calculate the tangential velocity of a rotating wheel with angular velocity of 32 rad/s with the wheel diameter of 30 cm. Given r = ½ (30cm) = 15 cm or 0.15 m. 𝜔 = 32 rad/s Solution 𝑣𝑡 = 𝜔 ∙ 𝑟 𝑟𝑎𝑑 𝑣𝑡 = (32 ) ∙ (0.15𝑚) 𝑠 𝑣𝑡 = 4.8 𝑚/𝑠 TANGENTIAL ACCELERATION The object moving in a circle doesn’t have any tangential acceleration or zero tangential acceleration it means that the object is moving with a constant velocity. When the object in circular motion changes the magnitudes and direction of the tangential velocity it resulted tangential acceleration. Tangential acceleration is a measure of how the tangential velocity of a point changes with time. Tangential acceleration is just like linear acceleration, but it’s particular to the tangential direction. It always acts perpendicular to the centripetal acceleration of the object moving in a circle. 𝑎𝑡 = 𝑎𝑡 = tangential acceleration dv = change in velocity dt = change in time 𝑑𝑣 𝑑𝑡 Example: A certain object accelerates uniformly in a circular path with a speed of 10 m/s to 100 m/s in 25 sec. Calculate the acceleration to tangential. Given: 𝑡𝑖 = 0 𝑠𝑒𝑐 – initial time Required: 𝑎𝑡 = ? 𝑡𝑓 = 25 𝑠𝑒𝑐 – final time 119 Specialized Subject - STEM 𝑣𝑖 = 10 𝑚/𝑠 – initial velocity 𝑣𝑓 = 100𝑚/𝑠 – final velocity Find the time taken 𝑑𝑡 = 𝑡𝑓 − 𝑡𝑖 = 30 sec − 0 sec = 25 𝑠𝑒𝑐𝑠 Find the change in velocity 𝑑𝑣 = 𝑣𝑓 − 𝑣𝑖 = 100 𝑚 𝑚 − 10 = 90 𝑚/𝑠 𝑠 𝑠 Find tangential acceleration 𝒂𝒕 = 𝑎𝑡 = 𝒅𝒗 𝒅𝒕 90 𝑚/𝑠 25 𝑠𝑒𝑐 𝑎𝑡 = 3.6 𝑚/𝑠 2 CENTRIPETAL ACCELERATION Uniform Circular Motion is the motion of the object in a circle with constant speed and as it moves in the circle it constantly changing in direction tangent to the path of the circle in any point. As it continuously changes in direction the velocity vector also changes and experienced acceleration. This acceleration is called as the centripetal acceleration, means center seeking. Fig. 1. The direction of the velocity vector of an object in uniform circular motion at any point is perpendicular to the centripetal force of the object. 120 Specialized Subject - STEM Fig. 2. Velocity Vector ∆𝑣 = 𝑣2 − 𝑣1 Where: ∆𝑣 – Change in velocity V1 and v2 – velocity in uniform circular motion For any object in uniform circular motion with a velocity in circular path with the radius, r. the magnitude of the centripetal acceleration is 𝑣2 𝑎𝑐 = 𝑟 Where: ac – centripetal acceleration v - velocity r – radius Example: A stone swings in a circle of radius 4 m. If its constant speed is 6 m/s, what is the centripetal acceleration? v = 6 m/s r=4m 𝑣2 𝑟 (6 𝑚/𝑠)2 𝑎𝑐 = 4𝑚 𝑎𝑐 = ac = 9 m/s2 Centripetal Force Centripetal force is the total force acting on the object in uniform circular motion and the direction is always towards the center of the rotation. In Newtons Second Law of Motion, the total force acting on the object causes the acceleration of mass, 𝐹𝑇𝑜𝑡𝑎𝑙 = 𝑚𝑎. Just like for the uniform circular motion the acceleration is the centripetal acceleration, a = ac. 121 Specialized Subject - STEM In Newtons Second Law of Motion equation 𝐹 = 𝑚𝑎 Where F – force m – mass a – acceleration 𝑎= 𝐹 𝑚 ---------------equation 1 Centripetal acceleration 𝑣2 𝑎𝑐 = 𝑟 ----------------equation 2 Where: ac – centripetal acceleration v - velocity r – radius Equate the equation 1 and 2 𝑎 = 𝑎𝑐 The magnitude of the centripetal force, Fc = mac Therefore, the Centripetal force, Fc in terms of tangential velocity is 𝐹𝑐 = 𝑚 𝑣2 𝑟 Where Fc – Centripetal Force m – mass v – velocity r – radius Example: 122 Specialized Subject - STEM Calculate the centripetal force exert on a 450 kg jeep taking a turn on a 330 m radius road at 20 m/s? Given mass – 450 kg radius – 330 m velocity – 20 m/s Solution 𝐹𝑐 = 𝑚 𝑣2 𝑟 (20 𝑚/𝑠)2 𝐹𝑐 = 450 𝑘𝑔 330 𝑚 𝐹𝑐 = 545 𝑘𝑔. 𝑚/𝑠 2 RADIUS OF CURVATURE The radius of curvature is defined as the radius of the approximate circle at a particular point. It is the length of the curvature vector. As the curve moves, the radius changes. It is denoted by r. Equation: 𝑚𝑣 2 𝑟𝑐 = 𝐹 rc = radius of curvature m = mass v = velocity F = lateral gripping force Example The minimum lift to a 900 kg helicopter is 9,000 N. if the helicopter travels at 110 m/s, calculate the possible radius of curvature. Use the centripetal force equation 𝐹 = 𝑚𝑣 2 𝐹 𝑚𝑣 2 𝑟𝑐 . Rearranging, we find that radius of curvature, 𝑟𝑐 = . Substitute the value minimum helicopter lift; 123 Specialized Subject - STEM 𝑚𝑣 2 𝑟𝑐 = 𝐹 (400 𝑘𝑔) (90 𝑚/𝑠)2 𝑟𝑐 = 9,000𝑁 𝑟𝑐 = 360 𝑚 UNDERSTANDING: Post-test I. Solve the following problems about circular motions. 1. What is the tangential acceleration of the object if it accelerates uniformly in circular motion with changes in velocity of 80 m/s an in the total changes of time of 25 seconds? 2. The ball is tied to a string to whirl it having a radius of 50 cm at a velocity of 1.2 m/s. What will be the acceleration of the ball? 3. What is the acceleration of the bicycle if the velocity is 1.5 m/s in a circular path with the radius of 75 meters? 4. A jeepney follows a circular road with a radius of 300 meters at a speed of 30 m/s. What is the magnitude of the jeepney’s acceleration? 5. A 450 kg jeep taking a turn on a 350 m radius road at 22 m/s. Calculate the centripetal force exerted on the jeep. Post Test Multiple Choice. Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper. 124 Specialized Subject - STEM 1. What is tangential velocity? a. Measured velocity of the object directed outward of the circle b. Measured velocity of the object at any point tangent to the circle c. Measured acceleration of the object at any point tangent to the circle d. Measured acceleration of the object directed outward of the circle 2. What is centripetal acceleration? 3. 4. 5. 6. 7. a. The acceleration of the object in non-uniform circular motion directed outward the circle b. The acceleration of the object in non-uniform circular motion directed inward the circle c. The acceleration of the object in uniform circular motion directed parallel to the path of the circle. d. The acceleration in uniform circular motion directed tangent to the path of the circle Which of the following is defined as the radius of the circle in a certain point and was denoted by r? a. Radius of curvature b. Measure of radius c. radians d. curve Which of the following is the measure of how the tangential velocity of a point changes with time? a. centripetal acceleration b. tangential acceleration b. tangential velocity c. radius of curvature Which of the following is the total force acting on the object moving in uniform uniform circular motion and the direction is always towards the center of the rotation? a. centripetal force b. centrifugal force c. normal force d. gravitational force What is the acceleration of a 500-kg tricycle moving at 8 m/s takes a turn around a circle with a radius of 20 m? a. 0.4 m/s2 b. 4 m/s2 c. 0.5 m/s2 d. 5 m/s2 What is the total force of acting on the 450 kg-tricycle is the acceleration is 4m/s2? a. 1700 kg.m/s2 125 Specialized Subject - STEM b. 1800 N c. 1900 N d. 1950 N 8. What is the centripetal force exerted on a 500 kg jeep taking a turn on a 300 m radius road on horizontal ground at 20 m/s? a. 33 kg.m/s2 b. 34 N c. 667 N d. 670 kg.m/s2 9. Which of the following acts on an object in uniform circular motion with direction is always towards the center of the rotation? a. Tangential velocity b. Tangential acceleration c. Centripetal acceleration d. Centripetal force 10. What is the tangential velocity of a rotating wheel with the diameter of 75 cm and angular velocity of 48 rad/sec? a. 0.64 m/s b. 64 m/s c. 36 m/s d. 2,160 m/s 11. Which of the following proves that there is a total force acting on an object in uniform circular motion? a. Total force is acting on the object accelerating b. There is total force because there is no acceleration c. No total force when there is acceleration d. Total force acts only in gravitational force 12. What is the acceleration of the ball tied to a string to whirl at the radius of 40 cm with the velocity of 1.1 m/s? a. 0.3 m/s2 b. 3.0 m/s2 c. 0.44 m/s2 d. 44 m/s2 13. What is the centripetal force exerted to a 400 kg jeep with a velocity of 25 m/s taking its turn to the curved road with a radius of 350 m? a. 29 kg.m/s2 b. 192 kg.m/s2 c. 30 N d. d.714 N 14. What is the possible radius of curvature of a 700kg helicopter with a minimum lift of 8,000 N that travels at 100 m/s. 126 Specialized Subject - STEM a. 8.75 m b. 76.56 m c. 875 m d. 1142.86 m 15. What is the tangential velocity of an object moving in a circle with the radius of 15 meters in constant angular velocity of 12 rad/s? a. 180 m/s b. 0.8 m/s c. 1.25 m/s d. 9.6 m/ 127 Specialized Subject - STEM MODULE3Q1 INERTIAL FRAME OF REFERENCE Lesson 1: Inertial Frame of Reference Overview: This module was designed and written with you in mind. It is here to help you master the accuracy and precision. The scope of this module permits it to be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. This module also included several tasks per each stage: priming, processing and understanding for students to work on to fully learn and master the competencies. Most Essential Learning Competencies: Define inertial frames of reference Lesson Learning Objectives: After going through this module, you are expected to: ✓ To know that any frame of reference in which the law of inertia is true is the inertial frame of reference. ✓ To know that a non-inertial frame is the accelerating object relative to inertial frame. ✓ To understand that the inertial frame of reference and all moving objects are relative PRE-TEST Direction: Read the questions properly, encircle the correct answer. 1. Newton’s first law of motion is also known as a. law of acceleration b. law of action and reaction c. law of inertia d. law of gravitation 2. Inertia is the property of mass in which an object at rest wants to stay at rest, and an object that is moving wants to _____. a. Also stay at rest. b. Stay moving in a straight line unless acted upon by another force. 128 Specialized Subject - STEM c. Stay moving in a circular motion unless acted upon by another force. d. Stay moving in a straight line, but only if it has been acted upon by another force. 3. Which of the following is an example of an inertial reference frame? a. a frame attached to an object on which there are no forces b. any reference frame that is at rest c. a reference frame attached to the center of the universe d. a reference frame attached to the Earth 4. Which of the following statements is not true for the inertial reference frame? a. a reference frame in which newton’s first law of motion is valid b. a reference frame in which the law of inertia holds true c. a reference frame which haves a constant increasing acceleration d. a reference frame which are not accelerating 5. In physics, frames of reference are classified by two main types: _____. a. true and fictional b. inertial and non-inertial c. fast and slow d. real and imagined PRIMING Two major types of forces 1. Contact Force 129 Specialized Subject - STEM 2. Noncontact Force PROCESSING Contact Force Contact force is a force that requires contact on both objects to occur. Contact forces are being everywhere and responsible for interactions applied between small and large objects. In Physics, contact force is the force acting at the point of contact between two objects against each other. Contact forces is subdivided into the following components, one is the force 130 Specialized Subject - STEM that is perpendicular to the surface of the object or the normal force, second is the force parallel to the surface of the object or the friction force, and forces that opposes fluids. Types of Contact Forces 1. Normal Force – a force exerted against the gravitational force present by the objects touching each other. Example of normal force a. the book is at rest on top of the table b. the box placed on the floor c. the eggs on the nest 2. Tensional Force- a force applied to a rope, string, or cable that makes them to be compressed or to be stretched by pulling on each side. Example of Tensional Force a. the pail was tied to the well b. the cradle was tied on the rope at two ends c. the star shaped Christmas lantern was hung on the ceiling 3. Frictional Force- a force created by both surfaces of the objects that is being rubbed against each other resulting by moving in either same direction or different direction. Example of Frictional Force a. the man is walking b. the girl slide to slides c. the boy rides to his bicycle 4. Air Resistance Force or Drag Force – is a force in the opposite direction of the object in air or fluid. Example of Air Resistance Force a. the sky diver jumps with his parachute b. dropping the paper from a 2-meter height c. the feather was flying through the air Noncontact Force Action at a Distance Forces is the other termed for noncontact forces and only results when two objects interact without any physical contact with each other. Regardless of their physical 131 Specialized Subject - STEM separation they can exert push or pull to the object. There are also different types of noncontact forces. Types of Noncontact Forces 1. Magnetic Force – attraction and repulsion resulted by putting together the end of same poles or different poles of the magnetic object. Magnetic force also resulted impacts of action induced by the electromagnetic materials to produced magnetic fields. Magnetic fields are surrounded and produced by magnetized material and by shifting into electrical charges such as those used in electromagnets. Example of Magnetic Force a. a compass b. ref magnets c. induction stove 2. Electrostatic Force-Just like magnetic forces, electrostatic force are either attractive or repulsive resulted by positive and negative charges of particles. Electrostatics force are resulted by like charges that repel like protons and unlike charges that attract like protons and electrons. Example of Electrostatic Force a. Combing hair with plastic comb b. rubbing the balloon in fur c. wiping of cloth into glass rod 3. Gravitational Force-is pulling of objects with masses towards the center of the earth. Example of Electrostatic Force a. ball dropped to the floor b. the boy riding his bicycle down the road c. The girl standing in top of the hill 132 Specialized Subject - STEM UNDERSTANDING: Post-test I. Tell whether the situation is in inertial or non- inertial frame of reference. Situation Inertial / Non- Inertial Reference Frame 1. The object at rest and in motion remains motion unless acted by a net force. 2. The object is accelerating either in linear fashion or rotating around some axis. 3. John is holding his a ball and riding on a bus that is moving with a constant velocity in a westward direction. 4. You are riding at the bus when suddenly the ball that you are holding falls down the floor of the bus. The bus starts to decelerate, then the ball on the floor accelerate forward inside the bus by itself. 5. Train moving with a constant velocity. 6. Car A is speeding up and passing car B. 7. A turning car with a constant velocity 8. Bea drop the stone from the third floor of a building. The stones falls down straight to the ground. 133 Specialized Subject - STEM 9. You and your friend is riding in a merry go round. You fells like not moving at all even the merry go round is continuously rotating to its center. 10. The driver is driving a vehicle moving at a constant speed at a straight road Post Test Multiple Choice. Choose the letter of the best answer. 1. Which of the following is a noncontact force? a. Drag Force b. Gravitational Force c. Tension Force d. Unbalanced Force 2. Which force is acting in the opposite direction of the object in motion? a. Tension b. Buoyant c. Friction d. Normal 3. What is the example of contact force? a. Rubbing your hands together b. picking paper clips by magnet c. falling stone d. putting near the two bar magnets 4. What happen when you put near together the different poles of the two bar magnets? a. reaction b. concentration c. repulsion d. attraction 5. Which of the following best describes the contact forces? a. forces between same objects b. forces between dissimilar objects c. forces between objects that touch d. forces between objects that do not touch 6. Which of the following is the force resulted by positive and negative charges of particles? a. magnetic force 134 Specialized Subject - STEM b. electrostatic force c. gravitational force d. frictional force 7. Which of the following is an example of gravitational force? a. combing hair with plastic comb b. wiping of cloth into the glass rod c. the cradle was tied on the rope at two ends d. ball dropped to the floor 8. Which of the following best describes the non-contact forces? a. forces between objects that do not touch b. forces between same objects c. forces between dissimilar objects d. forces between objects that touch 9. Which of the following is the force that pulls the objects with masses towards the center of the earth? a. normal force b. frictional force c. gravitational force d. air resistance force 10. Which of the following force is needed to apply to a string be stretched? a. Normal force b. frictional force c. drag force d. tensional force 11. Which of the following force the following is an example of normal force? a. book lifted up of the table b. book place at rest on top of the table c. book sliding at the table d. book falls at the edge of the table 12. Which of the following forces is a contact force? a. Air resistance force b. magnetic force c. electrostatic force d. gravitational force 13. What is another name of noncontact forces? a. length forces b. action at a distance force c. drag forces 135 Specialized Subject - STEM d. air resistance force 14. Which of the following force is parallel to the surface of the object? a. normal force b. drag force c. friction force d. tension force 15. Which of the following describe the force between two particles with the same charge? a. reaction b. concentration c. repulsion d. attraction 136