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Transcript
Physics Unit 1: Units of Measure & Basic Motion Review with Examples Standard Units of Measure used in Physics: SI Base Units of Measure Base Quantity Base Unit time Seconds distance/length meters mass kilograms temperature Kelvin electric current Amperes Abbreviation s m kg K A SI Derived Units of Measure Base Quantity Base Unit Force Newtons velocity/speed meters/second acceleration meters/second2 Abbreviation N m/s m/s2 SI Prefixes used in notation for simplifying written expressions: Prefix femto pico nano micro milli centi deci kilo mega giga tera Abbreviation f p n μ m c d k M G T SI Prefixes Decimal Scientific Notation 0.000000000000001 ×10 -15 0.000000000001 ×10 -12 0.000000001 ×10 -9 0.000001 ×10 -6 0.001 ×10 -3 0.01 ×10 -2 0.1 ×10 -1 1000 ×10 3 1000000 ×10 6 1000000000 ×10 9 1000000000000 ×10 12 Example femtometer (fs) picometer (pm) nanometer (nm) micrometer ( μ m) millimeter (mm) centimeter (cm) decimeter (dm) kilometer (km) megameter (Mm) gigameter (Gm) terameter (Tm) Conversions into standard units of measure: The magnitude of numbers can be written in many forms and many times utilize SI prefixes to shorten the written form. However, in order to work with these numbers more easily, they must be in similar units of measure. If they are not, then we must convert them into the standard base units necessary for that specific situation. Examples: 1.1) 75cm 75(.01)m 0.75m In this case the centi in centimeter is replaced by its decimal form (0.01)and multiplied to the original value (75). 1.2) 5GL 1.3) 8km / hr 5(1000000000) L 5000000000L 8(1000)m 2.22m / s 3600s In this case the kilo is replaced by its decimal form (1000) while the hours on the denominator are replaced with its value in seconds (3600). Scientific Notation: Scientific notation is a way of writing very large or small numbers to shorten their appearance and make them easier to compare with other numbers by utilizing a decimal rounded to no more than 2 decimal places multiplied by a ten with an exponent that represents the number of places the decimal would have to move to regenerate the original number. When writing the decimal, we must move the decimal in the original number (left or right) until there is only one digit (1,2,…7,8,9) on the left side of the decimal. The number of times the decimal was moved represents the number of tens that would need to be multiplied or divided into the original number in order to make the decimal place change positions. The total number of tens multiplied or divided into the original number then becomes the exponent for the notation form. If the decimal was moved to the right—the exponent is positive. If the decimal was moved to the left—the exponent is negative. Examples: 1.4) Here the decimal was moved 8 times to the right in order to get one digit (3) to the left side of the decimal. Thus there were 8 tens that would need to be multiplied to the decimal in order to get the original number, thus the exponent will be 8. 350,000,000 3.5 10 10 10 10 10 10 10 10 3.5 108 1.5) 0.00000000592 5.92 10 10 10 10 10 10 10 10 10 5.92 10 9 This time the decimal was moved 9 times to the right in order to get on digit (5) to the left side of the decimal. Thus there were 9 tens that must be divided into the decimal in order to get the original number. 3.95 1010 1.6) 39,452,087,000 1.7) 13379.2 109 1.34 104 109 1.34 105 1.8) (3.52 106 )(2.4 107 ) (3.52 2.4)(106107 ) 8.45 1013 Basics of Motion: Distance- Total measureable length of how far an object has moved or traveled Displacement- Shortest measureable distance between beginning and ending locations Speed- The rate at which an object changes positions or covers a distance Velocity- The speed at which an object is moving with an indicated direction Acceleration- The rate at which an object changes its velocity (speed up, slow down, change directions) v Examples: d t a v v v0 t t 1.9) A car travels 75 m in 13 s. Find its average velocity. d 75m t 13s v? v 1.10) A ball rolling at an average speed of 5 m/s rolls for 10 s. How far did the ball travel? v 5m / s t 10s d ? v 1.11) 1.12) A race car speeds up from 5 m/s to 32 m/s in 8.2 s. What is the average acceleration experienced by the car? A runner moving at 15 m/s attempts to stop after running passed the finish line by slowing with an average acceleration of 2.75m / s 2 . How long does it take him to come to a stop on the track? v 0 5m / s v 32m / s t 8. 2 s a? v0 15m / s d d 5m / s t 10s d (5m / s )(10s ) 50m v 32(m / s) 5(m / s) t 8.2 s 27m / s a 3.29m / s 2 8.2 s a a v 0m/ s a 2.75m / s 2 t ? d 75m 5.77m / s t 13s v t 2.75m / s 2 t 0m / s 15m / s t 15m / s 5.45s 2.75m / s 2 Graphing Motion: Dot Diagrams If we were to set up a camera and take pictures of a jogger running through the park each second of their jog for 5 seconds and then superimpose those 5 pictures onto on photograph, we could see the motion of the jogger in relation to their position as time passes. To simplify the picture we could replace the jogger with a dot. We would still see the same information. Examples: 1.13) Constant Velocity 1.14) Positive Acceleration (speeding up) 1.15) Negative Acceleration (slowing down) 1.16) All three forms represented Distance vs Time Graphs: Instead of the traditional x-y coordinate plane used in mathematics, physics uses a coordinate plane appropriate for the example using specific variables. In this case the vertical axis is the distance or position of the object in relation to the origin and the horizontal axis is the measure of time. As time passes, the distance increases or decreases, based on the slope of the line. Mathematically slope is described as “rise over run” (vertical change divided by horizontal change) and in the graph to the right the rise represents distance and the run represents time. This means that the “rise over run” is distance over time which is also known as velocity. This means that the slope of a distance vs time graph represents the velocity of the object. In this specific example the slope=5. With respect to physics the distance traveled was 30 m and the time it took was 6.0 s. Using both concepts, the velocity was 5 m/s. Four Types of Slope for Distance Graphs: Positive Slope or + velocity (moving forward) Negative Slope or - velocity (moving backwards) Zero Slope or 0 velocity (not moving) Undefined Slope or not possible Examples: 1.17) The graph to the right represents an object’s motion for 12 seconds of time. a) Describe the object’s motion for each segment. b) How fast was it moving during the first 4 seconds? c) How fast was it moving during the last 6 seconds d) How do you know the object was not moving from 4-6 s? e) What was the total distance traveled? f) What was the displacement of the object? a) The object moves forward 40 meters in 4 seconds before coming to a stop for the next 2 seconds. Finally it moves backward 20 meters during the last 6 seconds of the event. b) slope c) v rise dist ance d 40m velocity v 10m / s run time t 4s d 20m 3.33m / s t 6s d) There is no or zero slope for these two seconds. e) 40 m + 0 m + 20 m = 60 m f) 40 m + 0 m + (-20) m = 20 m 1.18) The graph to the right represents the motion of two neighbor’s commute home from work on the freeway. a) Who is traveling the fastest? b) Who has the farthest distance to travel to get home? c) What happens at the 45 second mark? d) What is the approximate velocity of each driver? a) Neighbor B (steeper slope) b) Neighbor B (starts 50 meters behind A) c) Neighbor B passes Neighbor A (190 meters) d) Neighbor A v Neighbor B v = d 190m 4.22m / s t 45s d (190 + 50) m = = 5.33m / s t 45 s Velocity vs Time Graphs: In this case the vertical axis is the velocity of the object and the horizontal axis is the measure of time. As time passes, the velocity increases or decreases, based on the slope of the line. Since slope is described as “rise over run” then the rise represents velocity and the run represents time. This means that the “rise over run” is velocity over time which is also known as acceleration. Therefore, the slope of a velocity vs time graph represents the acceleration of the object. In this specific example the slope=1.67. With respect to physics the velocity was 20 m/s and the time it took was 12.0 s. Using both concepts, the acceleration was 1.67 m/s. Furthermore, since the distance an object travels is defined by the velocity at which it travels, we can find the distance an object has moved by finding the area under the curve of the graph. In the above graph, the shape generated by the line is a triangle with a height of 20 and a base of 12. Therefore the area under the curve is half of 20 m/s multiplied by 12 s which is 12 bh 12 (12s)(20m / s) 120m Four Types of Slope for Velocity Graphs: Positive Slope or + acceleration (speeding up) Negative Slope or - acceleration (slowing down) Zero Slope or 0 acceleration (constant velocity) Undefined Slope or not possible Examples: 1.19) The graph to the left represents an object’s motion for 16 s of time. a) Describe the object’s motion for each segment. b) Find the average acceleration during the first 4 seconds? c) Find the average acceleration during the last 8 seconds d) How far did the car move during the first 4 seconds? e) What was the total distance traveled? a) For the first 4 seconds the car speeds up from 0m/s to 20 m/s. It then goes at a constant velocity for another 4 seconds before slowing to a stop over the last 8 seconds. b) slope rise changein velocity v 20m / s 0m / s acceleration a 5m / s 2 run time t 4s v 0m / s 20m / s 2.5m / s 2 t 8s d) area of a triangle= 12 bh 12 (4s)(20m / s) 40m c) a e) area of first triangle 1 2 + area of rectangle bh (4s)(20m / s) lw (4s)(20m / s) 1 2 + area of last triangle 1 2 bh 12 (8s)(20m / s) 40m 80m 80m 200m 1.20) The graph to the right represents 12 seconds of travel. a) Describe what Kyle & Jill do during the 12 seconds. b) What is the average acceleration for Kyle & Jill? c) What happens at the 6 second mark? d) What is the distance traveled by each person? a) Kyle is moving at a constant velocity of 20 m/s while Jill started from rest and speeds up to 40 m/s. v 20m / s 20m / s 0m / s 2 t 12 s v 40m / s 0m / s 3.33m / s 2 Jill= a t 12 s b) Kyle= a c) Both Jill and Kyle have the same velocity at this moment. d) Kyle (rectangle)= lw (12s)(20m / s) 240m & Jill (triangle)= 1 2 bh 12 (12s)(40m / s) 240m