Chapter 11 Motion

1. Chapter 11 Motion

2. Section 11-2 Motion

3. II. Speed and Velocity A. Speed Measurements involve distance and time. : Speed describes how fast an object moves. : To find speed, you must measure 2 quantities: 1. Distance traveled by an object miles, meters, feet, etc 2. Time it takes to travel that distance seconds, minutes, hours, etc. : The units for speed are distance/time : m/s or km/h or mi/h

4. II. Speed and Velocity B. Constant Speed is the simplest type of motion. : Constant Speed: when an object covers equal distances in equal amounts of time. : If a race car has a constant speed of 96 m/s, the car travels 96 m every second.

5. II. Speed and Velocity C. Speed can be determined from a distance-time graph. : We can see the relationship between distance, time, and speed by plotting a distance-time graph. : Time is on the X-Axis and distance is on the Y-Axis. : Speed can also be found by calculating the slope of a line. slope = rise = d2 – d1 = speed run t2 – t1 distance time

6. II. Speed and Velocity D. Speed is Calculated as Distance Divided by Time. : Most objects do not move with constantspeed. : “Average” Speed is simply the distance covered by an Object divided by the time. : Speed = (distance) / (time) or.. s = d/t

7. II. Speed and Velocity :Example… Suppose a wheelchair racer finishes a 132m race in 18seconds. What is the speed? s = ? d = 132m t = 18s s = d s = 132 t 18 s = 7.3 m/s The racer’s average speed is 7.3 m/s. The racer’s speed may have been faster or slower at different intervals of the race.

8. II. Speed and Velocity E. Velocity describes both speed and direction. : Sometimes describing the speed of an object is not enough. You may also need to know the direction in which the object is moving. : Velocity describes both speed and direction. : Direction can be—North, South, East, West of some point, or specify the angle from a fixed line. It can also be positive or negative in the line of motion.

9. Example Problems • A glacier moved 89m per day down the valley. Find the glacier’s velocity in m/s. v = ? d = 89m t = 1day x 24h= 24hr 24hr x 3600 = 86,400s v = d v = 89 t 86400 v = 0.0010 m/s down the valley

10. Example Problems 2. Find the velocity in meters per second of a swimmer who swims exactly 110m toward the shore in 72s. v = ? d = 110m t = 72s v = d v = 110 t 72 v = 1.5 m/s towards the shore

11. Example Problems 3. Find the velocity in meters per second of a baseball thrown 38m from third to first base in 1.7s. v = ? d = 38m t = 1.7s v = d v = 38 t 1.7 v = 22.35 m/s from third to first base

12. Example Problems 4. Calculate the distance in meters a cyclist would travel in 5 hours at an average velocity of 12 km/h to the southwest. d = ? v = 12km/h / 3.6= 3.33m/s t = 5h x 3600 = 18,000s d = tv d = (18,000s)(3.33m/s) d = 59940m

13. Example Problems 5. Calculate the time in seconds an Olympic skier would take to finish a 2.6km race at an average velocity of 28m/s downhill. v = 28m/s d = 2.6km x 1000= 2600m t = ? t = d t = 2600 v 28 t = 92.86s

14. Section 11-3 Acceleration

15. II. Acceleration A. Acceleration is any change in velocity. : To find acceleration of an object moving in a straight line, we need to measure the object’s velocity at different times. : Acceleration can be calculated by dividing the change in the object’s velocity by the time in which the change occurs. : Units are in m/s2.

16. Acceleration = final velocity-initial velocity time or.. a = vf – vi t : What does an acceleration value tell you????? -If the acceleration has a greater value, the object is speeding up more rapidly. : When you press on the gas pedal in a car, you speed up and your acceleration is in the direction of the car’s motion (positive). : When you press on the brake pedal, your acceleration is opposite to the direction of motion (negative) and you slow down.

17. Examples: 1) Natalie accelerates her skateboard along a straight path from 0m/s to 4m/s in 2.5s. Find her acceleration. vi = vf = a = t = a = vf – vi t 0 m/s 4 m/s a = 4 – 0 2.5 ? a = 4 2.5 2.5 s a = 1.6 m/s2

18. Examples: 2) A turtle swimming in a straight line toward shore has a speed of 0.5m/s. After 4s, its speed is 0.80m/s. What is the turtle’s acceleration? vi = vf = a = t = a = vf – vi t 0.5 m/s 0.80 m/s a = 0.80 – 0.5 4 ? 4 s a = 0.3 4 a = 0.075 m/s2

19. 12 m/s Examples: 3) Find the acceleration of a subway train that slows down from 12m/s to 9.6m/s in 0.8s. vi = vf = a = t = a = vf – vi t 9.6 m/s ? a = 9.6 – 12 0.8 0.8 s a = – 2.4 0.8 a = -3 m/s2 negative because slowing down!

20. Examples: 4) Marisa’s car accelerates at a rate of 2.6m/s2. Calculate how long it takes her car to accelerate from 24.6 m/s to 26.8 m/s. vi = vf = a = t = 24.6 m/s t = vf – vi a 26.8 m/s 2.6 m/s2 t = 26.8 – 24.6 2.6 ? t = 2.2 2.6 t = 0.846 s

21. Examples: 5) A bicycle travels at a constant velocity of 4.5m/s, and then speeds up with an acceleration of 2.3 m/s2. Calculate the bicycle’s speed after accelerating for 5.0s. vi = vf = a = t = 4.5 m/s ? Vf = at + vi Vf = (2.3 x 5) + 4.5 2.3 m/s2 vf = 11.5 + 4.5 5.0 s vf = 16 m/s

22. III. Velocity-Time Plots : Just as the slope of a distance-time plot tells you the speed, the slope of a velocity-time plot will tell you the acceleration. : Slope = rise = v2 – v1 = acceleration run t2 – t1 : If there is a positive slope, the object is speeding up. : If there is a negative slope, the object is slowing down. : If the slope is zero (a horizontal line), the object is moving at a constant velocity (or at rest). : When the plot is a straight line, then they change in velocity (the acceleration) isconstant. This means that it changes by the same amount each second.

23. IV. Other useful relationships : Already know: 1) v = d 2) a = vf – vo t t : The v in equation 1) is average velocity, you can write it as 3) vf + vo = d 2 t : By combining 2) and 3), you get 4) 2ad = vf2 – vo2 : By combining 2) and 4), you get 5) d = vot + ½ at2

24. Put on your equation sheet These equations (after a little rearranging) are: no d equation vf - vo = at no a equation 2d = vf + vo t no t equation 2ad = vf2 – vo2 no vf equation d = vot + ½ at2

25. Examples: 1) A car is traveling at a speed of 16m/s for a time of 180s. How far has it traveled if the car accelerates at a rate of 0.5m/s2? vo = vf = a = t = d = no vf equation d = vot + ½ at2 16 m/s d = (16)(180) + ½ (0.5)(180)2 0.5 m/s2 d = 2880 + 8100 180 s d = 10,980m ?

26. Examples: 2) A car starts from rest and accelerates at a rate of 4.5m/s2. How long does it take it to reach a speed of 60m/s? vo = vf = a = t = d = no d equation vf - vo = at 0 m/s 60 m/s 60 - 0 = (4.5)t 4.5 m/s2 60 = 4.5t ? t = 13.3 s

27. Examples: 3) A truck skids to a stop at a rate of -0.9m/s2. How fast was it originally traveling if it traveled 100m while stopping? vo = vf = a = t = d = no t equation 2ad = vf2 – vo2 ? 0 m/s 2(-0.9)(100) = 02 – vo2 - 0.9 m/s2 -180 = 0 – vo2 100 m vo2 = 180 vo = 13.4 m/s

28. no a equation 2d = vf + vo t Examples: 4) A car can go from 0mi/h to 60mi/h in 5s. How far does it travel in this amount of time? vo = vf = a = t = d = 0 mi/h 60 mi/h 2d = 60 + 0 0.00139 5 s = 0.00139 h 2d = 60 0.00139 ? 2d = 0.083 d = 0.042 mi

29. Chapter Summary The average speedof an object is defined as the distance the object travels divided by the time of travel. The distance-timegraph of an object moving at constant speed is a straight line. The slope of the line is the object’s speed. The SI unit for speed is meters per second, (m/s). The velocityof an object consists of both its speed and the direction of motion.

30. Acceleration is a change in thevelocityof an object. An object accelerates when it speeds up, slows down, or changes direction. Acceleration is caused by a force. For straight line motion, average acceleration as defined as the change in an object’s velocity per unit of time. The SI unit for acceleration is meters per second squared, (m/s2).