1. Motion Along a Straight Line
Chapter 2
Motion Along a Straight Line

2. Describe straight-line motion in terms of velocity and acceleration
Goals for Chapter 2
Describe straight-line motion in terms of velocity and acceleration
Distinguish between average and instantaneous velocity and acceleration
Interpret graphs
position versus time, x(t) versus t
(slope = velocity!)

3. Describe straight-line motion in terms of velocity and acceleration
Goals for Chapter 2
Describe straight-line motion in terms of velocity and acceleration
Distinguish between average and instantaneous velocity and acceleration
Interpret graphs
velocity versus time, v(t) versus t
Slope = acceleration!

4. Understand straight-line motion with constant acceleration
Goals for Chapter 2
Understand straight-line motion with constant acceleration
Examine freely falling bodies
Analyze straight-line motion when the acceleration is not constant 4

5. Introduction Kinematics is the study of motion.
Displacement, velocity and acceleration are important physical quantities.
A bungee jumper speeds up during the first part of his fall and then slows to a halt.

6. Displacement vs. Distance
Displacement (blue line) = how far the object is from its starting point, regardless of path
Distance traveled (dashed line) is measured along the actual path.
Figure 2-4. Caption: A person walks 70m east, then 30 m west. The total distance traveled is 100 m (path is shown dashed in black); but the displacement, shown as a solid blue arrow, is 40 m to the east. 6

7. Displacement vs. Distance
Q: You make a round trip to the store 1 mile away.
What distance do you travel?
What is your displacement?
Figure 2-4. Caption: A person walks 70m east, then 30 m west. The total distance traveled is 100 m (path is shown dashed in black); but the displacement, shown as a solid blue arrow, is 40 m to the east. 7

8. Displacement vs. Distance
Q: You walk 70 meters across the campus, hear a friend call from behind, and walk 30 meters back the way you came to meet her.
What distance do you travel?
What is your displacement?
Figure 2-4. Caption: A person walks 70m east, then 30 m west. The total distance traveled is 100 m (path is shown dashed in black); but the displacement, shown as a solid blue arrow, is 40 m to the east. 8

9. Displacement vs. Distance
Displacement is written:
SIGN matters! Direction matters!
It is a VECTOR!!
Displacement is negative =>
Figure 2-5. Caption: The arrow represents the displacement x2 – x1. Distances are in meters.
Figure 2-6. Caption: For the displacement Δx = x2 – x1 = 10.0 m – 30.0 m, the displacement vector points to the left.
<= Positive displacement 9

10. Speed vs. Velocity
Speed is how far an object travels in a given time interval (in any direction)
Ex: Go 10 miles to Chabot in 30 minutes
Average speed = 10 mi / 0.5 hr = 20 mph 10

11. Speed vs. Velocity Velocity includes directional information:
VECTOR!
Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes
Average velocity = 20 mi / hr = 60 mph NORTH 11

12. Speed vs. Velocity
Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes
Average velocity = 20 mi / hr = 60 mph NORTH 12

13. Speed vs. Velocity Velocity includes directional information:
VECTOR!
Ex: Go 10 miles on 880 Northbound to Chabot in 30 minutes
Average velocity = 10 mi / 0.5 hr = 20 mph NORTH 13

14. Speed vs. Velocity Speed is a SCALAR
60 miles/hour, 88 ft/sec, 27 meters/sec
Velocity is a VECTOR
60 mph North
88 ft/sec East
27 azimuth of 173 degrees 14

15. Example of Average Velocity
Position of runner as a function of time is plotted as moving along the x axis of a coordinate system.
During a 3.00-s time interval, a runner’s position changes from x1 = 50.0 m to x2 = 30.5 m
What was the runner’s average velocity?
Figure 2-7. Caption: Example 2–1.
A person runs from x1 = 50.0 m to x2 = 30.5 m. The displacement is –19.5 m.
Answer: Divide the displacement by the elapsed time; average velocity is m/s 15

16. Note! Dx = FINAL – INITIAL position
Example of Average Velocity
During a 3.00-sec interval, runner’s position changes from x1 = 50.0 m to x2 = 30.5 m What was the runner’s average velocity?
Vavg = ( ) meters/3.00 sec
= -6.5 m/s in the x direction.
The answer must have value1, units2, & DIRECTION3
Note! Dx = FINAL – INITIAL position
Figure 2-7. Caption: Example 2–1.
A person runs from x1 = 50.0 m to x2 = 30.5 m. The displacement is –19.5 m.
Answer: Divide the displacement by the elapsed time; average velocity is m/s 16

17. The answer must have value & units
Example of Average SPEED
During a 3.00-s time interval, the runner’s position changes from x1 = 50.0 m to x2 = 30.5 m. What was the runner’s average speed?
Savg = | | meters/3.00 sec = 6.5 m/s
Figure 2-7. Caption: Example 2–1.
A person runs from x1 = 50.0 m to x2 = 30.5 m. The displacement is –19.5 m.
Answer: Divide the displacement by the elapsed time; average velocity is m/s
The answer must have value & units 17

18. Negative velocity???
Average x-velocity is negative during a time interval if particle moves in negative x-direction for that time interval. 18

19. Displacement, time, and average velocity
A racing car starts from rest, and after 1 second is 19 meters from the starting line.
After the next 3 seconds, the car is 277 meters from the starting line.
What was its average velocity in those 3 seconds? 19

20. Displacement, time, and average velocity—Figure 2.1
Q A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds?
Solution Method:
What do you know? What do you need to find? What are the units? What might be a reasonable estimate?
DRAW it! Visualize what is happening. Create a coordinate system, label the drawing with everything.
Find what you need from what you know 20

21. Displacement, time, and average velocity—Figure 2.1
A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds?
“starts from rest” = initial velocity = 0
car moves along straight (say along an x-axis)
has coordinate x = 0 at t=0 seconds
has coordinate x=+19 meters at t =1 second
Has coordinate x=+277 meters at t = 1+3 = 4 seconds.

22. Displacement, time, and average velocity—Figure 2.1
Q A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average velocity in those 3 seconds? 22

23. A position-time graph—Figure 2.3
A position-time graph (an “x-t” graph) shows the particle’s position x as a function of time t.
Average x-velocity is related to the slope of an x-t graph.

24. Instantaneous Speed
Instantaneous speed is the average speed in the limit as the time interval becomes infinitesimally short.
Ideally, a speedometer would measure instantaneous speed; in fact, it measures average speed, but over a very short time interval. Note: It doesn’t measure direction!
Figure 2-8. Caption: Car speedometer showing mi/h in white, and km/h in orange. 24

25. Instantaneous Speed
Instantaneous velocity is the average velocity in the limit as the time interval becomes infinitesimally short.
Velocity is a vector; you must include direction! V = 27 m/s west…
Figure 2-8. Caption: Car speedometer showing mi/h in white, and km/h in orange. 25

26. Instantaneous velocity—Figure 2.4
The instantaneous velocity is the velocity at a specific instant of time or specific point along the path and is given by vx = dx/dt.

27. Instantaneous velocity—Figure 2.4
The instantaneous velocity is the velocity at a specific instant of time or specific point along the path and is given by vx = dx/dt.
The average speed is not the magnitude of the average velocity! 27

28. Instantaneous Velocity Example
A jet engine moves along an experimental track (the x axis) as shown.
Its position as a function of time is given by the equation
x = At2 + B
where A = 2.10 m/s2 and B = 2.80 m.
Figure Caption: Example 2–3. (a) Engine traveling on a straight track. (b) Graph of x vs. t: x = At2 + B.
Solution: a. Using the equation given, at 3.00 s, the engine is at 21.7 m; at 5.00 s it is at 55.3 m, so the displacement is 33.6 m.
b. The average velocity is the displacement divided by the time: 16.8 m/s.
c. Take the derivative: v = dx/dt = 2At = 21.0 m/s. 28

29. Instantaneous Velocity Example
A jet engine’s position as a function of time is x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m.
Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s.
Determine the average velocity during this time interval.
Determine the magnitude of the instantaneous velocity at t = 5.00 s.
Figure Caption: Example 2–3. (a) Engine traveling on a straight track. (b) Graph of x vs. t: x = At2 + B.
Solution: a. Using the equation given, at 3.00 s, the engine is at 21.7 m; at 5.00 s it is at 55.3 m, so the displacement is 33.6 m.
b. The average velocity is the displacement divided by the time: 16.8 m/s.
c. Take the derivative: v = dx/dt = 2At = 21.0 m/s. 29

30. Instantaneous Velocity Example
A jet engine’s position as a function of time is x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m.
Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s.
@ t = 3.00 s x1 = 21.7
@ t = 5.00 s x2 = 55.3
x2 – x1 = 33.6 meters in +x direction
Figure Caption: Example 2–3. (a) Engine traveling on a straight track. (b) Graph of x vs. t: x = At2 + B.
Solution: a. Using the equation given, at 3.00 s, the engine is at 21.7 m; at 5.00 s it is at 55.3 m, so the displacement is 33.6 m.
b. The average velocity is the displacement divided by the time: 16.8 m/s.
c. Take the derivative: v = dx/dt = 2At = 21.0 m/s. 30

31. Instantaneous Velocity Example
A jet engine’s position as a function of time is x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m.
b) Determine the average velocity during this time interval.
Vavg = 33.6 m/ 2.00 sec
= 16.8 m/s
in the + x direction
Figure Caption: Example 2–3. (a) Engine traveling on a straight track. (b) Graph of x vs. t: x = At2 + B.
Solution: a. Using the equation given, at 3.00 s, the engine is at 21.7 m; at 5.00 s it is at 55.3 m, so the displacement is 33.6 m.
b. The average velocity is the displacement divided by the time: 16.8 m/s.
c. Take the derivative: v = dx/dt = 2At = 21.0 m/s. 31

32. Instantaneous Velocity Example
A jet engine’s position as a function of time is x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m.
(c) Determine the magnitude of the instantaneous velocity at t = 5.00 s
|v| = t = 5.00 seconds
= 21.0 m/s
Figure Caption: Example 2–3. (a) Engine traveling on a straight track. (b) Graph of x vs. t: x = At2 + B.
Solution: a. Using the equation given, at 3.00 s, the engine is at 21.7 m; at 5.00 s it is at 55.3 m, so the displacement is 33.6 m.
b. The average velocity is the displacement divided by the time: 16.8 m/s.
c. Take the derivative: v = dx/dt = 2At = 21.0 m/s. 32

33. Acceleration = the rate of change of velocity.
Units: meters/sec/sec or m/s^2 or m/s2 or ft/s2
Since velocity is a vector, acceleration is ALSO a vector, so direction is crucial…
A = 2.10 m/s2 in the +x direction
Figure Caption: Example 2–4.The car is shown at the start with v1 = 0 at t1 = 0. The car is shown three more times, at t = 1.0 s, t = 2.0 s, and at the end of our time interval, t2 = 5.0 s. We assume the acceleration is constant and equals 5.0 m/s2. The green arrows represent the velocity vectors; the length of each arrow represents the magnitude of the velocity at that moment. The acceleration vector is the orange arrow. Distances are not to scale.
Solution: The average acceleration is the change in speed divided by the time, 5.0 m/s2. 33

34. A car accelerates along a straight road from rest to 90 km/h in 5.0 s.
Acceleration Example
A car accelerates along a straight road from rest to 90 km/h in 5.0 s.
What is the magnitude of its average acceleration?
KEY WORDS:
“straight road” = assume constant acceleration
“from rest” = starts at 0 km/h
Figure Caption: Example 2–4.The car is shown at the start with v1 = 0 at t1 = 0. The car is shown three more times, at t = 1.0 s, t = 2.0 s, and at the end of our time interval, t2 = 5.0 s. We assume the acceleration is constant and equals 5.0 m/s2. The green arrows represent the velocity vectors; the length of each arrow represents the magnitude of the velocity at that moment. The acceleration vector is the orange arrow. Distances are not to scale.
Solution: The average acceleration is the change in speed divided by the time, 5.0 m/s2. 34

35. |a| = 5.0 m/s2 (note – magnitude only is requested)
Acceleration Example
A car accelerates along a straight road from rest to 90 km/h in 5.0 s
What is the magnitude of its average acceleration?
|a| = (90 km/hr – 0 km/hr)/5.0 sec
= 18 km/h/sec along road
better – convert to more reasonable units
90 km/hr = 90 x 103 m/hr x 1hr/3600 s = 25 m/s So
|a| = 5.0 m/s2 (note – magnitude only is requested)
Figure Caption: Example 2–4.The car is shown at the start with v1 = 0 at t1 = 0. The car is shown three more times, at t = 1.0 s, t = 2.0 s, and at the end of our time interval, t2 = 5.0 s. We assume the acceleration is constant and equals 5.0 m/s2. The green arrows represent the velocity vectors; the length of each arrow represents the magnitude of the velocity at that moment. The acceleration vector is the orange arrow. Distances are not to scale.
Solution: The average acceleration is the change in speed divided by the time, 5.0 m/s2. 35

36. Acceleration Acceleration = the rate of change of velocity.
Figure Caption: Example 2–4.The car is shown at the start with v1 = 0 at t1 = 0. The car is shown three more times, at t = 1.0 s, t = 2.0 s, and at the end of our time interval, t2 = 5.0 s. We assume the acceleration is constant and equals 5.0 m/s2. The green arrows represent the velocity vectors; the length of each arrow represents the magnitude of the velocity at that moment. The acceleration vector is the orange arrow. Distances are not to scale.
Solution: The average acceleration is the change in speed divided by the time, 5.0 m/s2. 36

37. Acceleration vs. Velocity?
If the velocity of an object is zero, does it mean that the acceleration is zero?
(b) If the acceleration is zero, does it mean that the velocity is zero? Think of some examples.
Solution: a. No; if this were true nothing could ever change from a velocity of zero!
b. No, but it does mean the velocity is constant. 37

38. Acceleration Example
An automobile is moving to the right along a straight highway. Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to slow down to v2 = 5.0 m/s, what was the car’s average acceleration?
Figure Caption: Example 2–6, showing the position of the car at times t1 and t2, as well as the car’s velocity represented by the green arrows. The acceleration vector (orange) points to the left as the car slows down while moving to the right.
Solution: The average acceleration is the change in speed divided by the time; it is negative because it is in the negative x direction, and the car is slowing down: a = -2.0 m/s2 38

39. Acceleration Example
An automobile is moving to the right along a straight highway. Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to slow down to v2 = 5.0 m/s, what was the car’s average acceleration?
Figure Caption: Example 2–6, showing the position of the car at times t1 and t2, as well as the car’s velocity represented by the green arrows. The acceleration vector (orange) points to the left as the car slows down while moving to the right.
Solution: The average acceleration is the change in speed divided by the time; it is negative because it is in the negative x direction, and the car is slowing down: a = -2.0 m/s2 39

40. A semantic difference between negative acceleration and deceleration:
Acceleration Example
A semantic difference between negative acceleration and deceleration:
“Negative” acceleration is acceleration in the negative direction (defined by coordinate system).
“Deceleration” occurs when the acceleration is opposite in direction to the velocity.
Figure Caption: The car of Example 2–6, now moving to the left and decelerating. The acceleration is +2.0 m/s. 40

41. Finding velocity on an x-t graph
At any point on an x-t graph, the instantaneous x-velocity is equal to the slope of the tangent to the curve at that point.

42. Motion diagrams
A motion diagram shows position of a particle at various instants, and arrows represent its velocity at each instant.

43. Average acceleration
Acceleration describes the rate of change of velocity with time.
The average x-acceleration is aav-x = vx/t.

44. Instantaneous acceleration
The instantaneous acceleration is ax = dvx/dt.
Follow Example 2.3, which illustrates an accelerating racing car.

45. Average Acceleration
Q A racing car starts from rest, and after 1 second is 19 meters from the starting line. After the next 3 seconds, the car is 277 meters from the starting line. What was its average acceleration in the first second?
What was its average acceleration in the first 4 seconds? 45

46. Finding acceleration on a vx-t graph
As shown in Figure 2.12, the x-t graph may be used to find the instantaneous acceleration and the average acceleration.

47. A vx-t graph and a motion diagram
Figure 2.13 shows the vx-t graph and the motion diagram for a particle.

48. An x-t graph and a motion diagram
Figure 2.14 shows the x-t graph and the motion diagram for a particle.

49. Motion with constant acceleration—Figures 2.15 and 2.17
For a particle with constant acceleration, the velocity changes at the same rate throughout the motion.

50. Acceleration given x(t)
A particle is moving in a straight line with its position is given by x = (2.10 m/s2)t2 + (2.80 m). Calculate (a) its average acceleration during the interval from t1 = 3.00 s to t2 = 5.00 s, & (b) its instantaneous acceleration as a function of time.
Figure Caption: Example 2–7. Graphs of (a) x vs. t, (b) v vs. t, and (c) a vs. t for the motion x = At2 + B. Note that increases linearly with and that the acceleration a is constant. Also, v is the slope of the x vs. t curve, whereas a is the slope of the v vs. t curve.
Solution: The velocity at time t is the derivative of x; v = (4.20 m/s2)t.
a. Solve for v at the two times; a = 4.20 m/s2.
b. Take the derivative of v: a = 4.20 m/s2. 50

51. Acceleration given x(t)
A particle is moving in a straight line with its position is given by x = (2.10 m/s2)t2 + (2.80 m). Calculate (a) its average acceleration during the interval from t1 = 3.00 s to t2 = 5.00 s
V = dx/dt = (4.2 m/s) t
V1 = 12.6 m/s
V2 = 21 m/s
Dv/Dt = 8.4 m/s/2.0 s = 4.2 m/s2
Figure Caption: Example 2–7. Graphs of (a) x vs. t, (b) v vs. t, and (c) a vs. t for the motion x = At2 + B. Note that increases linearly with and that the acceleration a is constant. Also, v is the slope of the x vs. t curve, whereas a is the slope of the v vs. t curve.
Solution: The velocity at time t is the derivative of x; v = (4.20 m/s2)t.
a. Solve for v at the two times; a = 4.20 m/s2.
b. Take the derivative of v: a = 4.20 m/s2. 51

52. Acceleration given x(t)
A particle is moving in a straight line with its position is given by x = (2.10 m/s2)t2 + (2.80 m). Calculate (b) its instantaneous acceleration as a function of time.
Figure Caption: Example 2–7. Graphs of (a) x vs. t, (b) v vs. t, and (c) a vs. t for the motion x = At2 + B. Note that increases linearly with and that the acceleration a is constant. Also, v is the slope of the x vs. t curve, whereas a is the slope of the v vs. t curve.
Solution: The velocity at time t is the derivative of x; v = (4.20 m/s2)t.
a. Solve for v at the two times; a = 4.20 m/s2.
b. Take the derivative of v: a = 4.20 m/s2. 52

53. Analyzing acceleration
Graph shows Velocity as a function of time for two cars accelerating from 0 to 100 km/h in a time of 10.0 s Compare (a) the average acceleration; (b) instantaneous acceleration; and (c) total distance traveled for the two cars.
Figure 2-19.
Solution: a. Average acceleration is the same; both have the same change in speed over the same time.
b. Car A accelerates faster than B at the beginning but then slower than B towards the end (look at the slope of the lines).
c. Car A is always going faster than car B, so it will travel farther. 53

54. Analyzing acceleration
Velocity as a function of time for two cars accelerating from 0 to 100 km/h in a time of 10.0 s Compare (a) the average acceleration; (b) instantaneous acceleration; and (c) total distance traveled for the two cars.
Same final speed in time =>
Same average acceleration
But Car A accelerates faster…
Figure 2-19.
Solution: a. Average acceleration is the same; both have the same change in speed over the same time.
b. Car A accelerates faster than B at the beginning but then slower than B towards the end (look at the slope of the lines).
c. Car A is always going faster than car B, so it will travel farther. 54

55. Constant Acceleration Equations
FIVE key variables:
Dxdisplacement vinitial , vfinal , acceleration time
FIVE key equations:
Dx = ½ (vi+vf)t
Dx = vit + ½ at2
Dx = vft – ½ at2
vf = vi + at
vf2 = vi2 + 2aDx

56. The equations of motion with constant acceleration
Equation of Motion Variables Present
Initial velocity, final velocity, acceleration, time
Displacement (x – x0), initial velocity, time, acceleration
Initial velocity, final velocity, acceleration, displacement
Displacement, initial velocity, final velocity, time
Displacement (x – x0), final velocity, time, acceleration
x = x0 + vxt – 1/2axt2 56

57. The equations of motion with constant acceleration
Equation of Motion Find 3 of 4, solve for 4th!
Initial velocity, final velocity, acceleration, time
Displacement (x – x0), initial velocity, time, acceleration
Initial velocity, final velocity, acceleration, displacement
Displacement, initial velocity, final velocity, time
Initial velocity, final velocity, acceleration, time
Displacement (x – x0), initial velocity, time, acceleration
Initial velocity, final velocity, acceleration, displacement
Displacement, initial velocity, final velocity, time
Displacement (x – x0), final velocity, time, acceleration
x = x0 + vxt – 1/2axt2 57

58. A motorcycle with constant acceleration
Follow Example 2.4 for an accelerating motorcycle.

59. Two bodies with different accelerations
Two different initial/final velocities and accelerations
TIME and displacement are linked!

60. Freely falling bodies
Free fall is the motion of an object under the influence of only gravity.
In the figure, a strobe light flashes with equal time intervals between flashes.
The velocity change is the same in each time interval, so the acceleration is constant.

61. Freely Falling Objects
In the absence of air resistance, all objects fall with the same acceleration, although this may be tricky to tell by testing in an environment where there is air resistance.
Figure Caption: (a) A ball and a light piece of paper are dropped at the same time. (b) Repeated, with the paper wadded up. 61

62. Freely Falling Objects
The acceleration due to gravity at the Earth’s surface is approximately 9.80 m/s2. At a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration.
Figure Caption: A rock and a feather are dropped simultaneously (a) in air, (b) in a vacuum. 62

63. Freely Falling Objects
Example: Falling from a tower.
Suppose that a ball is dropped (v0 = 0) from a tower 70.0 m high. How far will it have fallen after a time t1 = 1.00 s, t2 = 2.00 s, and t3 = 3.00 s? Ignore air resistance.
Figure Caption: Example 2–14. (a) An object dropped from a tower falls with progressively greater speed and covers greater distance with each successive second. (See also Fig. 2–26.) (b) Graph of y vs. t.
Solution: We are given the acceleration, the initial speed, and the time; we need to find the distance. Substituting gives t1 = 4.90 m, t2 = 19.6 m, and t3 = 44.1 m. 63

64. Example: Thrown down from a tower
Suppose a ball is thrown downward with an initial velocity of 3.00 m/s, instead of being dropped. (a) What then would be its position after 1.00 s and 2.00 s? (b) What is its speed after 1.00 s and 2.00 s? Compare with the speeds of a dropped ball.
Solution: This is the same as Example 2-14, except that the initial speed is not zero.
At t = 1.00 s, y = 7.90 m. At t = 2.00 s, y = 25.6 m.
At t = 1.00 s, v = 12.8 m/s. At t = 2.00 s, v = 22.6 m/s. The speed is always 3.00 m/s faster than a dropped ball. 64

65. Freely Falling Objects
Example: Ball thrown up!
A person throws a ball upward into the air with an initial velocity of 15.0 m/s
Calculate (a) how high it goes, & (b) how long the ball is in the air before it comes back to the hand. Ignore air resistance.
Figure Caption: An object thrown into the air leaves the thrower’s hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2–16, 2–17, 2–18, and 2–19.
Solution: a. At the highest position, the speed is zero, so we know the acceleration, the initial and final speeds, and are asked for the distance. Substituting gives y = 11.5 m.
b. Now we want the time; t = 3.06 s. 65

66. Up-and-down motion in free fall
An object is in free fall even when it is moving upward.

67. Is the acceleration zero at the highest point?—Figure 2.25
The vertical velocity, but not the acceleration, is zero at the highest point.

68. Ball thrown upward (cont.)
Consider again a ball thrown upward, & calculate (a) how much time it takes for the ball to reach the maximum height, (b) the velocity of the ball when it returns to the thrower’s hand (point C).
The time is 1.53 s, half the time for a round trip (since we are ignoring air resistance).
v = m/s 68

69. Freely Falling Objects
Give examples to show the error in these two common misconceptions: (1) that acceleration and velocity are always in the same direction (2) that an object thrown upward has zero acceleration at the highest point.
If acceleration and velocity were always in the same direction, nothing could ever slow down!
At its highest point, the speed of thrown object is zero. If its acceleration were also zero, it would just stay at that point. 69

70. The quadratic formula
For a ball thrown upward at an initial speed of 15.0 m/s, calculate at what time t the ball passes a point 8.00 m above the person’s hand.
Figure Caption: Graphs of (a) y vs. t, (b) v vs. t for a ball thrown upward, Examples 2–16, 2–18, and 2–19.
Solution: We are given the initial and final position, the initial speed, and the acceleration, and want to find the time. This is a quadratic equation; there are two solutions: t = 0.69 s and t = 2.37 s. The first is the ball going up and the second is the ball coming back down. 70

71. Ball thrown at edge of cliff
A ball is thrown upward at a speed of 15.0 m/s by a person on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below. (a) How long does it take the ball to reach the base of the cliff? (b) What is the total distance traveled by the ball? Ignore air resistance (likely to be significant, so our result is an approximation).
Figure Caption: Example 2–20. The person in Fig. 2–30 stands on the edge of a cliff. The ball falls to the base of the cliff, 50.0 m below.
Solution: a. We use the same quadratic formula as before, we find t = 5.07 s (the negative solution is physically meaningless).
b. The ball goes up 11.5 m, then down 11.5 m + 50 m, for a total distance of 73.0 m. 71

72. Variable Acceleration; Integral Calculus
Deriving the kinematic equations through integration:
For constant acceleration, 72

73. Variable Acceleration; Integral Calculus
Then:
For constant acceleration, 73

74. Variable Acceleration; Integral Calculus
Example: Integrating a time-varying acceleration.
An experimental vehicle starts from rest (v0 = 0) at t = 0 and accelerates at a rate given by a = (7.00 m/s3)t. What is
its velocity and
its displacement 2.00 s later?
Solution: a. Integrate to find v = (3.50 m/s3)t2 = 14.0 m/s.
b. Integrate again to find x = (3.50 m/s3)t3/3 = 9.33 m. 74

75. Graphical Analysis and Numerical Integration
The total displacement of an object can be described as the area under the v-t curve: 75

76. Graphical Analysis and Numerical Integration
Similarly, the velocity may be written as the area under the a-t curve.
However, if the velocity or acceleration is not integrable, or is known only graphically, numerical integration may be used instead. 76

77. Example: Numerical integration
An object starts from rest at t = 0 and accelerates at a rate a(t) = (8.00 m/s4)t2. Determine its velocity after 2.00 s using numerical methods.
Figure 2-35.
Solution: The figure illustrates the process. Using the given intervals, v = 21.0 m/s, compared to the calculated value of m/s. 77

78. Velocity and position by integration
The acceleration of a car is not always constant.
The motion may be integrated over many small time intervals to give

79. Motion with changing acceleration
Follow Example 2.9.
Figure 2.29 illustrates the motion graphically.