Document related concepts
no text concepts found
Transcript
```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
Kilo
K
103
Kilometer
Km
103 m
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
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
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
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
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.
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
b. one fewer significant digit
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
c. precision in measurement
d. significant digits in measurement
9. Random error lead to a lack of:
a. accuracy in measurement
Specialized Subject - STEM
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
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
-
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
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
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
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
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).
vectors-and-scalars-0
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.
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.
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
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
The vector and its components form a right angled triangle as shown below.
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
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
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
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
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
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
For numbers 1-2
The following graph represents the position as a function of time for a moving object. Use this
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.
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
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
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
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
velocity – 20 m/s
Solution
𝐹𝑐 = 𝑚
𝑣2
𝑟
(20 𝑚/𝑠)2
𝐹𝑐 = 450 𝑘𝑔
330 𝑚
𝐹𝑐 = 545 𝑘𝑔. 𝑚/𝑠 2
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
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
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
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?
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
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
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
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
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?
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
```
Related documents