Download presentation

Presentation is loading. Please wait.

1
**Chabot College Physics 4A Spring 2015 Chapter 2**

Motion Along a Straight Line!

2
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!) time Position x(t)

3
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! time Velocity v(t)

4
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
**Kinematics is the study of motion.**

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
Introduction A bungee jumper speeds up during the first part of his fall and then slows to a halt. Model his position in time with function x(t) Model his velocity in time v(t) Model his acceleration in time a(t)

7
**2-1 Position, Displacement, and Average Velocity**

For this chapter, we restrict motion in three ways: We consider motion along a straight line only We discuss only the motion itself, not the forces that cause it We consider the moving object to be a particle © 2014 John Wiley & Sons, Inc. All rights reserved.

8
**2-1 Position, Displacement, and Average Velocity**

A particle is either: A point-like object (such as an electron) Or an object that moves such that each part travels in the same direction at the same rate (no rotation or stretching) © 2014 John Wiley & Sons, Inc. All rights reserved.

9
**2-1 Position, Displacement, and Average Velocity**

Position is measured relative to a reference point: The origin, or zero point, of an axis x at time t1 Origin x = 0 © 2014 John Wiley & Sons, Inc. All rights reserved.

10
**2-1 Position, Displacement, and Average Velocity**

Position is still measured relative to the same reference point: The origin, or zero point, of an axis x at time t2 Origin x = 0 © 2014 John Wiley & Sons, Inc. All rights reserved.

11
**2-1 Position, Displacement, and Average Velocity**

Position has a sign: Positive direction is in the direction of increasing numbers Negative direction is opposite the positive © 2014 John Wiley & Sons, Inc. All rights reserved.

12
**2-1 Position, Displacement, and Average Velocity**

A change in position is called displacement ∆x is the change in x Always (final position) – (initial position) © 2014 John Wiley & Sons, Inc. All rights reserved.

13
**Displacement Displacement is written: SIGN matters! Direction matters!**

It is a VECTOR!! Vectors have THREE things… Magnitude Direction Units 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. Displacement could be: 3 m North, or – 5 cm on the x axis, or miles NW 13

14
**Displacement 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 14

15
**2-1 Position, Displacement, and Average Velocity**

Examples A particle moves . . . From x = 5 m to x = 12 m? ∆x = 7 m (positive direction) © 2014 John Wiley & Sons, Inc. All rights reserved.

16
**2-1 Position, Displacement, and Average Velocity**

Examples A particle moves . . . From x = 5 m to x = 12 m: ∆x = 7 m (positive direction) From x = 5 m to x = 1 m? ∆x = -4 m (negative direction) © 2014 John Wiley & Sons, Inc. All rights reserved.

17
**2-1 Position, Displacement, and Average Velocity**

Examples A particle moves . . . From x = 5 m to x = 12 m: ∆x = 7 m (positive direction) From x = 5 m to x = 1 m: ∆x = -4 m (negative direction) From x = 5 m to x = 200 m to x = 5 m: ∆x = 0 m !! The actual distance covered is irrelevant © 2014 John Wiley & Sons, Inc. All rights reserved.

18
**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. 18

19
**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. 19

20
**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. 20

21
**2-1 Position, Displacement, and Average Velocity**

Answer: pairs (b) and (c) (b) -7 m – -3 m = -4 m (c) -3 m – 7 m = -10 m © 2014 John Wiley & Sons, Inc. All rights reserved.

22
Speed vs. Velocity Speed is the ratio of 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 22

23
**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 Always think about your answer …. Is it reasonable?? 23

24
Speed vs. Velocity Not with traffic is 60 mph even possible! Ex: Go 20 miles on 880 Northbound to Chabot in 20 minutes Average velocity = 20 mi / hr = 60 mph NORTH 24

25
**2-1 Position, Displacement, and Average Velocity**

Average velocity is the ratio of: A displacement, ∆x To the time interval in which the displacement occurred, ∆t Average velocity has units of (distance) / (time) Meters per second, m/s © 2014 John Wiley & Sons, Inc. All rights reserved.

26
**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 26

27
**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 27

28
**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 28

29
**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 29

30
**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 but it is a scalar! No direction needed 30

31
Negative velocity??? Average x-velocity is negative during a time interval if particle moves in negative x-direction for that time interval. 31

32
**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? 32

33
**Displacement, time, and average velocity**

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 33

34
**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? “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.

35
**Displacement, time, and average velocity**

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? 35

36
**A position-time graph x(t) vs t**

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.

37
**A position-time graph x(t) vs t**

38
**2-2 Instantaneous Velocity and Speed**

Example The graph shows the position and velocity of an elevator cab over time. The slope of x(t), and so also the velocity v, is zero from 0 to 1 s, and from 9s on. During the interval bc, the slope is constant and nonzero, so the cab moves with constant velocity (4 m/s).

39
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. 39

40
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. 40

41
**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. X(t) 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. time 41

42
**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. 42

43
**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. 43

44
**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. 44

45
**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. 45

46
**2-2 Instantaneous Velocity and Speed**

© 2014 John Wiley & Sons, Inc. All rights reserved.

47
**2-2 Instantaneous Velocity and Speed**

Answers: (a) Situations 1 and 4 (zero) (b) Situations 2 and 3 © 2014 John Wiley & Sons, Inc. All rights reserved.

48
**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… acceleration = 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. 48

49
**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. 49

50
**|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. 50

51
**Acceleration Acceleration = the rate of change of velocity. 51**

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. 51

52
**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. 52

53
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 53

54
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 54

55
**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. 55

56
**Finding velocity on an x-t graph**

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

57
Motion diagrams A motion diagram shows position of a particle at various instants, and arrows represent its velocity at each instant.

58
Motion diagrams A motion diagram shows position of a particle at various instants, and arrows represent its velocity at each instant.

59
Average acceleration Acceleration describes the rate of change of velocity with time. The average x-acceleration is aav-x = vx/t.

60
**Finding acceleration on a vx-t graph**

Use x vs. t graph to find instantaneous acceleration and average acceleration. Finding acceleration on a vx-t graph

61
**Instantaneous acceleration**

The instantaneous acceleration is ax = dvx/dt. 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?

62
**An x-t graph and a motion diagram**

63
**An x-t graph and a motion diagram**

64
**A vx-t graph and motion diagram**

65
**A vx-t graph and motion diagram**

66
**Motion with constant acceleration**

For a particle with constant acceleration, the velocity changes at the same rate throughout the motion.

67
**Acceleration given x(t)**

A particle is moving in a straight line with position 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) instantaneous acceleration as 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. 67

68
**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. 68

69
**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. 69

70
**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. 70

71
**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. 71

72
**2-4 Constant Acceleration**

In many cases acceleration is constant, or nearly so. For these cases, 5 special equations can be used. Note that constant acceleration means a velocity with a constant slope, and a position with varying slope (unless a = 0).

73
**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

74
**2-4 Constant Acceleration**

First basic equation When the acceleration is constant, the average and instantaneous accelerations are equal Rewrite Eq. 2-7 and rearrange Eq. (2-11) This equation reduces to v = v0 for t = 0 Its derivative yields the definition of a, dv/dt © 2014 John Wiley & Sons, Inc. All rights reserved.

75
**2-4 Constant Acceleration**

Second basic equation Eq. (2-12) Average = ((initial) + (final)) / 2: Eq. (2-13) Eq. (2-14) Substitute 2-14 into 2-12 Eq. (2-15) © 2014 John Wiley & Sons, Inc. All rights reserved.

76
**2-4 Constant Acceleration**

These are enough to solve any constant acceleration problem Solve as simultaneous equations Additional useful forms: Eq. (2-16) Eq. (2-17) Eq. (2-18) © 2014 John Wiley & Sons, Inc. All rights reserved.

77
**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 77

78
**The equations of motion with constant acceleration**

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 x = x0 + vxt – 1/2axt2 Equation of Motion Find 3 of 4, solve for 4th! 78

79
**2-4 Constant Acceleration**

Table 2-1 shows 5 key equations & the quantities missing from each.

80
**2-4 Constant Acceleration**

Answer: Situations 1 (a = 0) and 4. © 2014 John Wiley & Sons, Inc. All rights reserved.

81
**A motorcycle with constant acceleration**

Follow Example 2.4 for an accelerating motorcycle.

82
**Two bodies with different (constant) accelerations**

A motorist is moving at a constant speed of 15m/s (in a 25 mph zone, through a school crosswalk) passes a police officer on a motorcycle, who is stopped. The officer immediately accelerates at a constant 3.0 m/s2 in pursuit, and overtakes the motorist. How far are they from the point where the motorist first passed the officer? How much time elapsed?

83
**Two bodies with different accelerations**

A motorist is moving at a constant speed of 15m/s (in a 25 mph zone, through a school crosswalk) passes a police officer on a motorcycle, who is stopped. The officer immediately accelerates at a constant 3.0 m/s2 in pursuit, and overtakes the motorist.

84
**Two bodies with different accelerations**

Two different initial/final velocities and accelerations TIME and displacement are linked!

85
**Two bodies with different accelerations**

Two different initial/final velocities and accelerations TIME and displacement are linked!

86
**Two bodies with different accelerations (part 2!)**

tp tm A motorist is moving at a constant speed of 15m/s (in a 25 mph zone, through a school crosswalk) passes a police officer on a motorcycle, who is stopped. The officer waits 3 seconds, and then accelerates at a constant 3.0 m/s2 in pursuit, and overtakes the motorist.

87
**Two bodies with different accelerations**

tm Officer tp Displacement (m) Motorist Time (s) Two different initial/final velocities and accelerations & times? TIME and displacement are STILL linked, but times aren’t equal…

88
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.

89
**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. 89

90
**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. 90

91
**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. 91

92
**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. 92

93
**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 t = 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. 93

94
**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. 94

95
**Freely Falling Objects**

Example: Ball thrown up! A person standing on top of a building 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. 95

96
**Up-and-down motion in free fall**

An object is in free fall even when it is moving upward.

97
**Is the acceleration zero at the highest point?**

The vertical velocity, but not the acceleration, is zero at the highest point.

98
**Is the acceleration zero at the highest point?**

The vertical velocity, but not the acceleration, is zero at the highest point.

99
**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 99

100
**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. 100

101
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. 101

102
**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. 102

103
**Variable Acceleration; Integral Calculus**

Deriving the kinematic equations through integration: For constant acceleration, 103

104
**Variable Acceleration; Integral Calculus**

Then: For constant acceleration, 104

105
**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. 105

106
**Graphical Analysis and Numerical Integration**

The total displacement of an object can be described as the area under the v-t curve: 106

107
**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. 107

108
**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. 108

109
**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

110
**Motion with changing acceleration**

111
**2-3 Acceleration Acceleration is a vector quantity:**

Positive sign means in the positive coordinate direction Negative sign means the opposite Units of (distance) / (time squared)

112
**© 2014 John Wiley & Sons, Inc. All rights reserved.**

2-3 Acceleration Note: accelerations can be expressed in units of g Eq. (2-10) © 2014 John Wiley & Sons, Inc. All rights reserved.

113
**© 2014 John Wiley & Sons, Inc. All rights reserved.**

2-3 Acceleration Example If a car with velocity v = -25 m/s is braked to a stop in 5.0 s, then a = m/s Acceleration is positive, but speed has decreased. © 2014 John Wiley & Sons, Inc. All rights reserved.

114
**© 2014 John Wiley & Sons, Inc. All rights reserved.**

2-3 Acceleration © 2014 John Wiley & Sons, Inc. All rights reserved.

115
**© 2014 John Wiley & Sons, Inc. All rights reserved.**

2-3 Acceleration Answers: (a) + (b) - (c) - (d) + © 2014 John Wiley & Sons, Inc. All rights reserved.

116
**2-3 Acceleration Example**

The graph shows the velocity and acceleration of an elevator cab over time. When acceleration is 0 (e.g. interval bc) velocity is constant. When acceleration is positive (ab) upward velocity increases. When acceleration is negative (cd) upward velocity decreases. Steeper slope of the velocity-time graph indicates a larger magnitude of acceleration: the cab stops in half the time it takes to get up to speed.

117
**2-5 Free-Fall Acceleration**

Learning Objectives 2.16 Identify that if a particle is in free flight (whether upward or downward) and if we can neglect the effects of air on its motion, the particle has a constant downward acceleration with a magnitude g that we take to be 9.8m/s2. 2.17 Apply the constant acceleration equations (Table 2-1) to free-fall motion. Figure 2-12 © 2014 John Wiley & Sons, Inc. All rights reserved.

118
**2-5 Free-Fall Acceleration**

Free-fall acceleration is the rate at which an object accelerates downward in the absence of air resistance Varies with latitude and elevation Written as g, standard value of 9.8 m/s2 Independent of the properties of the object (mass, density, shape, see Figure 2-12) The equations of motion in Table 2-1 apply to objects in free-fall near Earth's surface In vertical flight (along the y axis) Where air resistance can be neglected © 2014 John Wiley & Sons, Inc. All rights reserved.

119
**2-5 Free-Fall Acceleration**

The free-fall acceleration is downward (-y direction) Value -g in the constant acceleration equations Answers: (a) The sign is positive (the ball moves upward); (b) The sign is negative (the ball moves downward); (c) The ball's acceleration is always -9.8 m/s2 at all points along its trajectory © 2014 John Wiley & Sons, Inc. All rights reserved.

120
**2-6 Graphical Integration in Motion Analysis**

2.19 Determine a particle's change in position by graphical integration on a graph of velocity versus time. 2.18 Determine a particle's change in velocity by graphical integration on a graph of acceleration versus time. © 2014 John Wiley & Sons, Inc. All rights reserved.

121
**2-6 Graphical Integration in Motion Analysis**

Integrating acceleration: Given a graph of an object's acceleration a versus time t, we can integrate to find velocity The Fundamental Theorem of Calculus gives: Eq. (2-27) © 2014 John Wiley & Sons, Inc. All rights reserved.

122
**2-6 Graphical Integration in Motion Analysis**

The definite integral can be evaluated from a graph: Eq. (2-28) © 2014 John Wiley & Sons, Inc. All rights reserved.

123
**2-6 Graphical Integration in Motion Analysis**

Example The graph shows the acceleration of a person's head and torso in a whiplash incident. To calculate the torso speed at t = s (assuming an initial speed of 0), find the area under the pink curve: area A = 0 area B = 0.5 (0.060 s) (50 m/s2) = 1.5 m/s area C = (0.010 s) (50 m/s2) = 0.50 m/s total area = 2.0 m/s © 2014 John Wiley & Sons, Inc. All rights reserved.

124
**© 2014 John Wiley & Sons, Inc. All rights reserved.**

2 Summary Position Relative to origin Positive and negative directions Displacement Change in position (vector) Eq. (2-1) Average Velocity Displacement / time (vector) Average Speed Distance traveled / time Eq. (2-2) Eq. (2-3) © 2014 John Wiley & Sons, Inc. All rights reserved.

125
**© 2014 John Wiley & Sons, Inc. All rights reserved.**

2 Summary Instantaneous Velocity At a moment in time Speed is its magnitude Average Acceleration Ratio of change in velocity to change in time Eq. (2-7) Eq. (2-4) Instantaneous Acceleration First derivative of velocity Second derivative of position Constant Acceleration Includes free-fall, where a = -g along the vertical axis Eq. (2-8) Tab. (2-1) © 2014 John Wiley & Sons, Inc. All rights reserved.

Similar presentations

OK

Chapter 2 Motion in One Dimension. Kinematics Describes motion while ignoring the external agents that might have caused or modified the motion For now,

Chapter 2 Motion in One Dimension. Kinematics Describes motion while ignoring the external agents that might have caused or modified the motion For now,

© 2018 SlidePlayer.com Inc.

All rights reserved.

To ensure the functioning of the site, we use **cookies**. We share information about your activities on the site with our partners and Google partners: social networks and companies engaged in advertising and web analytics. For more information, see the Privacy Policy and Google Privacy & Terms.
Your consent to our cookies if you continue to use this website.

Ads by Google

Ppt on indian politics swami Ppt on management by objectives drucker Ppt on pin diode phase Ppt on brand marketing salary Ppt on sbi mutual fund Ppt on producers consumers and decomposers in the amazon Ppt on payroll management system Ppt on appropriate climate responsive technologies for inclusive growth and sustainable development Viewer ppt online Ppt on operating system and its types