Lecture Presentation Chapter 2 Motion in One Dimension
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1 Lecture Presentation Chapter 2 Motion in One Dimension
2 Suggested Videos for Chapter 2 Prelecture Videos Motion Along a Line Acceleration Free Fall Video Tutor Solutions Motion in One Dimension Class Videos Motion Along a Straight Line Motion with Constant Acceleration Free Fall Slide 2-2
3 Suggested Simulations for Chapter 2 ActivPhysics PhETs The Moving Man Equation Grapher Slide 2-3
4 Chapter 2 Motion in One Dimension Chapter Goal: To describe and analyze linear motion. Slide 2-4
5 Chapter 2 Preview Looking Ahead: Uniform Motion Successive images of the rider are the same distance apart, so the velocity is constant. This is uniform motion. You ll learn to describe motion in terms of quantities such as distance and velocity, an important first step in analyzing motion. Slide 2-5
6 Chapter 2 Preview Looking Ahead: Acceleration A cheetah is capable of very high speeds but, more importantly, it is capable of a rapid change in speed a large acceleration. You ll use the concept of acceleration to solve problems of changing velocity, such as races, or predators chasing prey. Slide 2-6
7 Chapter 2 Preview Looking Ahead: Free Fall When you toss a coin, the motion both going up and coming down is determined by gravity alone. We call this free fall. How long does it take the coin to go up and come back down? This is the type of free-fall problem you ll learn to solve. Slide 2-7
8 Chapter 2 Preview Looking Ahead Text: p. 28 Slide 2-8
9 Chapter 2 Preview Looking Back: Motion Diagrams As you saw in Section 1.5, a good first step in analyzing motion is to draw a motion diagram, marking the position of an object in subsequent times. In this chapter, you ll learn to create motion diagrams for different types of motion along a line. Drawing pictures like this is a good staring point for solving problems. Slide 2-9
10 Chapter 2 Preview Stop to Think A bicycle is moving to the left with increasing speed. Which of the following motion diagrams illustrates this motion? Slide 2-10
11 Section 2.1 Describing Motion
12 Representing Position We will use an x-axis to analyze horizontal motion and motion on a ramp, with the positive end to the right. We will use a y-axis to analyze vertical motion, with the positive end up. Slide 2-12
13 Representing Position The motion diagram of a student walking to school and a coordinate axis for making measurements Every dot in the motion diagram of Figure 2.2 represents the student s position at a particular time. Figure 2.3 shows the student s motion shows the student s position as a graph of x versus t. Slide 2-13
14 QuickCheck 2.1 Here is a motion diagram of a car moving along a straight road: Which position-versus-time graph matches this motion diagram? Slide 2-14
15 QuickCheck 2.1 Here is a motion diagram of a car moving along a straight road: Which position-versus-time graph matches this motion diagram? E. Slide 2-15
16 From Position to Velocity On a position-versus-time graph, a faster speed corresponds to a steeper slope. The slope of an object s position-versus-time graph is the object s velocity at that point in the motion. Slide 2-16
17 From Position to Velocity Text: p. 31 Slide 2-17
18 From Position to Velocity We can deduce the velocity-versus-time graph from the positionversus-time graph. The velocity-versus-time graph is yet another way to represent an object s motion. Slide 2-18
19 Slide 2-19
20 Slide 2-20
21 Slide 2-21
22 QuickCheck 2.2 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? E. None of the above. Slide 2-22
23 QuickCheck 2.2 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? C. E. None of the above. Slide 2-23
24 QuickCheck 2.3 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? Slide 2-24
25 QuickCheck 2.3 Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? D. Slide 2-25
26 QuickCheck 2.4 A graph of position versus time for a basketball player moving down the court appears as follows: Which of the following velocity graphs matches the position graph? A. B. C. D. Slide 2-26
27 QuickCheck 2.4 A graph of position versus time for a basketball player moving down the court appears as follows: Which of the following velocity graphs matches the position graph? A. B. C. D. Slide 2-27
28 Example 2.2 Analyzing a car s position graph FIGURE 2.11 gives the position-versus-time graph of a car. a. Draw the car s velocityversus-time graph. b. Describe the car s motion in words. Slide 2-28
29 Example 2.2 Analyzing a car s position graph FIGURE 2.11 gives the position-versus-time graph of a car. a. Draw the car s velocityversus-time graph. b. Describe the car s motion in words. PREPARE Figure 2.11 is a graphical representation of the motion. The car s position-versus-time graph is a sequence of three straight lines. Each of these straight lines represents uniform motion at a constant velocity. We can determine the car s velocity during each interval of time by measuring the slope of the line. Slide 2-29
30 Example 2.2 Analyzing a car s position graph (cont.) SOLVE a. From t = 0 s to t = 2 s (Δt = 2 s) the car s displacement is Δx = 4 m 0 m = 4 m. The velocity during this interval is The car s position does not change from t = 2 s to t = 4 s (Δx = 0 m), so v x = 0 m/s. Finally, the displacement between t = 4 s and t = 6 s (Δt = 2 s) is Δx = 10 m. Thus the velocity during this interval is These velocities are represented graphically in FIGURE Slide 2-30
31 Example 2.2 Analyzing a car s position graph (cont.) SOLVE b. The velocity-versus-time graph of Figure 2.12 shows the motion in a way that we can describe in a straightforward manner: The car backs up for 2 s at 2 m/s, sits at rest for 2 s, then drives forward at 5 m/s for 2 s. ASSESS Notice that the velocity graph and the position graph look completely different. They should! The value of the velocity graph at any instant of time equals the slope of the position graph. Since the position graph is made up of segments of constant slope, the velocity graph should be made up of segments of constant value, as it is. This gives us confidence that the graph we have drawn is correct. Slide 2-31
32 From Velocity to Position We can deduce the position-versus-time graph from the velocity-versustime graph. The sign of the velocity tells us whether the slope of the position graph is positive or negative. The magnitude of the velocity tells us how steep the slope is. Slide 2-32
33 QuickCheck 2.6 A graph of velocity versus time for a hockey puck shot into a goal appears as follows: Which of the following position graphs matches the velocity graph? A. B. C. D. Slide 2-33
34 QuickCheck 2.6 A graph of velocity versus time for a hockey puck shot into a goal appears as follows: Which of the following position graphs matches the velocity graph? A. B. C. (d) D. Slide 2-34
35 QuickCheck 2.7 Which velocity-versus-time graph goes with this position graph? Slide 2-35
36 QuickCheck 2.7 Which velocity-versus-time graph goes with this position graph? C. Slide 2-36
37 Section 2.2 Uniform Motion
38 Reading Question 2.1 The slope at a point on a position-versus-time graph of an object is the Object s speed at that point. Object s average velocity at that point. Object s instantaneous velocity at that point. Object s acceleration at that point. Distance traveled by the object to that point. Slide 2-38
39 Reading Question 2.1 The slope at a point on a position-versus-time graph of an object is the Object s speed at that point. Object s average velocity at that point. Object s instantaneous velocity at that point. Object s acceleration at that point. Distance traveled by the object to that point. Slide 2-39
40 Uniform Motion Straight-line motion in which equal displacements occur during any successive equal-time intervals is called uniform motion or constantvelocity motion. An object s motion is uniform if and only if its position-versus-time graph is a straight line. Slide 2-40
41 Equations of Uniform Motion The velocity of an object in uniform motion tells us the amount by which its position changes during each second. Slide 2-41
42 Equations of Uniform Motion The velocity of an object in uniform motion tells us the amount by which its position changes during each second. The displacement Δx is proportional to the time interval Δt. Slide 2-42
43 Equations of Uniform Motion Text: p. 34 Slide 2-43
44 QuickCheck 2.8 Here is a position graph of an object: At t = 1.5 s, the object s velocity is A. 40 m/s B. 20 m/s C. 10 m/s D. 10 m/s E. None of the above Slide 2-44
45 QuickCheck 2.8 Here is a position graph of an object: At t = 1.5 s, the object s velocity is A. 40 m/s B. 20 m/s C. 10 m/s D. 10 m/s E. None of the above Slide 2-45
46 Example 2.3 If a train leaves Cleveland at 2:00 A train is moving due west at a constant speed. A passenger notes that it takes 10 minutes to travel 12 km. How long will it take the train to travel 60 km? Slide 2-46
47 Example 2.3 If a train leaves Cleveland at 2:00 A train is moving due west at a constant speed. A passenger notes that it takes 10 minutes to travel 12 km. How long will it take the train to travel 60 km? PREPARE For an object in uniform motion, Equation 2.5 shows that the distance traveled Δx is proportional to the time interval Δt, so this is a good problem to solve using ratio reasoning. Slide 2-47
48 Example 2.3 If a train leaves Cleveland at 2:00 (cont.) SOLVE We are comparing two cases: the time to travel 12 km and the time to travel 60 km. Because Δx is proportional to Δt, the ratio of the times will be equal to the ratio of the distances. The ratio of the distances is This is equal to the ratio of the times: It takes 10 minutes to travel 12 km, so it will take 50 minutes 5 times as long to travel 60 km. Slide 2-48
49 Slide 2-49
50 Slide 2-50
51 From Velocity to Position, One More Time The displacement Δx is equal to the area under the velocity graph during the time interval Δt. Slide 2-51
52 QuickCheck 2.11 Here is the velocity graph of an object that is at the origin (x = 0 m) at t = 0 s. At t = 4.0 s, the object s position is A. 20 m B. 16 m C. 12 m D. 8 m E. 4 m Slide 2-52
53 QuickCheck 2.11 Here is the velocity graph of an object that is at the origin (x = 0 m) at t = 0 s. At t = 4.0 s, the object s position is A. 20 m B. 16 m C. 12 m D. 8 m E. 4 m Displacement = area under the curve Slide 2-53
54 Section 2.3 Instantaneous Velocity
55 Reading Question 2.2 Which of the following is an example of uniform motion? A. A car going around a circular track at a constant speed. B. A person at rest starts running in a straight line in a fixed direction. C. A ball dropped from the top of a building. D. A hockey puck sliding in a straight line at a constant speed. Slide 2-55
56 Reading Question 2.2 Which of the following is an example of uniform motion? A. A car going around a circular track at a constant speed. B. A person at rest starts running in a straight line in a fixed direction. C. A ball dropped from the top of a building. D. A hockey puck sliding in a straight line at a constant speed. Slide 2-56
57 Instantaneous Velocity For one-dimensional motion, an object changing its velocity is either speeding up or slowing down. An object s velocity a speed and a direction at a specific instant of time t is called the object s instantaneous velocity. From now on, the word velocity will always mean instantaneous velocity. Slide 2-57
58 Finding the Instantaneous Velocity If the velocity changes, the position graph is a curved line. But we can compute a slope at a point by considering a small segment of the graph. Let s look at the motion in a very small time interval right around t = 0.75 s. This is highlighted with a circle, and we show a closeup in the next graph. Slide 2-58
59 Finding the Instantaneous Velocity In this magnified segment of the position graph, the curve isn t apparent. It appears to be a line segment. We can find the slope by calculating the rise over the run, just as before: v x = (1.6 m)/(0.20 s) = 8.0 m/s This is the slope at t = 0.75 s and thus the velocity at this instant of time. Slide 2-59
60 Finding the Instantaneous Velocity Graphically, the slope of the curve at a point is the same as the slope of a straight line drawn tangent to the curve at that point. Calculating rise over run for the tangent line, we get v x = (8.0 m)/(1.0 s) = 8.0 m/s This is the same value we obtained from the closeup view. The slope of the tangent line is the instantaneous velocity at that instant of time. Slide 2-60
61 Instantaneous Velocity Even when the speed varies we can still use the velocityversus-time graph to determine displacement. The area under the curve in a velocity-versus-time graph equals the displacement even for non-uniform motion. Slide 2-61
62 QuickCheck 2.5 The slope at a point on a position-versus-time graph of an object is A. The object s speed at that point. B. The object s velocity at that point. C. The object s acceleration at that point. D. The distance traveled by the object to that point. E. I am not sure. Slide 2-62
63 QuickCheck 2.5 The slope at a point on a position-versus-time graph of an object is A. The object s speed at that point. B. The object s velocity at that point. C. The object s acceleration at that point. D. The distance traveled by the object to that point. E. I am not sure. Slide 2-63
64 QuickCheck 2.9 When do objects 1 and 2 have the same velocity? A. At some instant before time t 0 B. At time t 0 C. At some instant after time t 0 D. Both A and B E. Never Slide 2-64
65 QuickCheck 2.9 When do objects 1 and 2 have the same velocity? A. At some instant before time t 0 B. At time t 0 C. At some instant after time t 0 D. Both A and B E. Never Same slope at this time Slide 2-65
66 QuickCheck 2.10 Masses P and Q move with the position graphs shown. Do P and Q ever have the same velocity? If so, at what time or times? A. P and Q have the same velocity at 2 s. B. P and Q have the same velocity at 1 s and 3 s. C. P and Q have the same velocity at 1 s, 2 s, and 3 s. D. P and Q never have the same velocity. Slide 2-66
67 QuickCheck 2.10 Masses P and Q move with the position graphs shown. Do P and Q ever have the same velocity? If so, at what time or times? A. P and Q have the same velocity at 2 s. B. P and Q have the same velocity at 1 s and 3 s. C. P and Q have the same velocity at 1 s, 2 s, and 3 s. D. P and Q never have the same velocity. Slide 2-67
68 Example 2.5 The displacement during a rapid start FIGURE 2.21 shows the velocity-versus-time graph of a car pulling away from a stop. How far does the car move during the first 3.0 s? Slide 2-68
69 Example 2.5 The displacement during a rapid start FIGURE 2.21 shows the velocity-versus-time graph of a car pulling away from a stop. How far does the car move during the first 3.0 s? PREPARE Figure 2.21 is a graphical representation of the motion. The question How far? indicates that we need to find a displacement Δx rather than a position x. According to Equation 2.7, the car s displacement Δx = x f x i between t = 0 s and t = 3 s is the area under the curve from t = 0 s to t = 3 s. Slide 2-69
70 Example 2.5 The displacement during a rapid start (cont.) SOLVE The curve in this case is an angled line, so the area is that of a triangle: The car moves 18 m during the first 3 seconds as its velocity changes from 0 to 12 m/s. Slide 2-70
71 Example 2.5 The displacement during a rapid start (cont.) ASSESS The physically meaningful area is a product of s and m/s, so Δx has the proper units of m. Let s check the numbers to see if they make physical sense. The final velocity, 12 m/s, is about 25 mph. Pulling away from a stop, you might expect to reach this speed in about 3 s at least if you have a reasonably sporty vehicle! If the car had moved at a constant 12 m/s (the final velocity) during these 3 s, the distance would be 36 m. The actual distance traveled during the 3 s is 18 m half of 36 m. This makes sense, as the velocity was 0 m/s at the start of the problem and increased steadily to 12 m/s. Slide 2-71
72 QuickCheck 2.13 A car moves along a straight stretch of road. The following graph shows the car s position as a function of time: At what point (or points) do the following conditions apply? The displacement is zero. The speed is zero. The speed is increasing. The speed is decreasing. Slide 2-72
73 QuickCheck 2.13 A car moves along a straight stretch of road. The following graph shows the car s position as a function of time: At what point (or points) do the following conditions apply? The displacement is zero. The speed is zero. The speed is increasing. The speed is decreasing. D B, E C A Slide 2-73
74 Section 2.4 Acceleration
75 Reading Question 2.3 The area under a velocity-versus-time graph of an object is A. The object s speed at that point. B. The object s acceleration at that point. C. The distance traveled by the object. D. The displacement of the object. E. This topic was not covered in this chapter. Slide 2-75
76 Reading Question 2.3 The area under a velocity-versus-time graph of an object is A. The object s speed at that point. B. The object s acceleration at that point. C. The distance traveled by the object. D. The displacement of the object. E. This topic was not covered in this chapter. Slide 2-76
77 Acceleration We define a new motion concept to describe an object whose velocity is changing. The ratio of Δv x /Δt is the rate of change of velocity. The ratio of Δv x /Δt is the slope of a velocity-versus-time graph. Slide 2-77
78 Units of Acceleration In our SI unit of velocity, 60 mph = 27 m/s. The Corvette speeds up to 27 m/s in Δt = 3.6 s. It is customary to abbreviate the acceleration units (m/s)/s as m/s 2, which we say as meters per second squared. Every second, the Corvette s velocity changes by 7.5 m/s. Slide 2-78
79 Slide 2-79
80 Example 2.6 Animal acceleration Lions, like most predators, are capable of very rapid starts. From rest, a lion can sustain an acceleration of 9.5 m/s 2 for up to one second. How much time does it take a lion to go from rest to a typical recreational runner s top speed of 10 mph? Slide 2-80
81 Example 2.6 Animal acceleration Lions, like most predators, are capable of very rapid starts. From rest, a lion can sustain an acceleration of 9.5 m/s 2 for up to one second. How much time does it take a lion to go from rest to a typical recreational runner s top speed of 10 mph? PREPARE We can start by converting to SI units. The speed the lion must reach is The lion can accelerate at 9.5 m/s 2, changing its speed by 9.5 m/s per second, for only 1.0 s long enough to reach 9.5 m/s. It will take the lion less than 1.0 s to reach 4.5 m/s, so we can use a x = 9.5 m/s 2 in our solution. Slide 2-81
82 Example 2.6 Animal acceleration (cont.) SOLVE We know the acceleration and the desired change in velocity, so we can rearrange Equation 2.8 to find the time: ASSESS The lion changes its speed by 9.5 meters per second in one second. So it s reasonable (if a bit intimidating) that it will reach 4.5 m/s in just under half a second. Slide 2-82
83 Representing Acceleration An object s acceleration is the slope of its velocityversus-time graph. Slide 2-83
84 Representing Acceleration We can find an acceleration graph from a velocity graph. Slide 2-84
85 QuickCheck 2.12 A particle has velocity as it moves from point 1 to point 2. The acceleration is shown. What is its velocity vector as it moves away from point 2? Slide 2-85
86 QuickCheck 2.12 A particle has velocity as it moves from point 1 to point 2. The acceleration is shown. What is its velocity vector as it moves away from point 2? B. Slide 2-86
87 QuickCheck 2.14 The motion diagram shows a particle that is slowing down. The sign of the position x and the sign of the velocity v x are: A. Position is positive, velocity is positive. B. Position is positive, velocity is negative. C. Position is negative, velocity is positive. D. Position is negative, velocity is negative. Slide 2-87
88 QuickCheck 2.14 The motion diagram shows a particle that is slowing down. The sign of the position x and the sign of the velocity v x are: A. Position is positive, velocity is positive. B. Position is positive, velocity is negative. C. Position is negative, velocity is positive. D. Position is negative, velocity is negative. Slide 2-88
89 Slide 2-89
90 The Sign of the Acceleration An object can move right or left (or up or down) while either speeding up or slowing down. Whether or not an object that is slowing down has a negative acceleration depends on the direction of motion. Slide 2-90
91 The Sign of the Acceleration (cont.) An object can move right or left (or up or down) while either speeding up or slowing down. Whether or not an object that is slowing down has a negative acceleration depends on the direction of motion. Slide 2-91
92 Slide 2-92
93 Slide 2-93
94 QuickCheck 2.15 The motion diagram shows a particle that is slowing down. The sign of the acceleration a x is: A. Acceleration is positive. B. Acceleration is negative. Slide 2-94
95 QuickCheck 2.15 The motion diagram shows a particle that is slowing down. The sign of the acceleration a x is: A. Acceleration is positive. B. Acceleration is negative. Slide 2-95
96 QuickCheck 2.16 A cyclist riding at 20 mph sees a stop sign and actually comes to a complete stop in 4 s. He then, in 6 s, returns to a speed of 15 mph. Which is his motion diagram? Slide 2-96
97 QuickCheck 2.16 A cyclist riding at 20 mph sees a stop sign and actually comes to a complete stop in 4 s. He then, in 6 s, returns to a speed of 15 mph. Which is his motion diagram? B. Slide 2-97
98 QuickCheck 2.17 These four motion diagrams show the motion of a particle along the x-axis. 1. Which motion diagrams correspond to a positive acceleration? 2. Which motion diagrams correspond to a negative acceleration? Slide 2-98
99 QuickCheck 2.17 These four motion diagrams show the motion of a particle along the x-axis. 1. Which motion diagrams correspond to a positive acceleration? 2. Which motion diagrams correspond to a negative acceleration? Positive Negative Positive Negative Slide 2-99
100 QuickCheck 2.18 Gustav jumps out of a tree and lands on a trampoline. The trampoline sags 2 feet before launching Gustav back into the air. At the very bottom, where the sag is the greatest, Gustav s acceleration is A. Upward. B. Downward. C. Zero. Slide 2-100
101 QuickCheck 2.18 Gustav jumps out of a tree and lands on a trampoline. The trampoline sags 2 feet before launching Gustav back into the air. At the very bottom, where the sag is the greatest, Gustav s acceleration is A. Upward. B. Downward. C. Zero. Slide 2-101
102 QuickCheck 2.19 A cart slows down while moving away from the origin. What do the position and velocity graphs look like? Slide 2-102
103 QuickCheck 2.19 A cart slows down while moving away from the origin. What do the position and velocity graphs look like? D. Slide 2-103
104 QuickCheck 2.20 A cart speeds up toward the origin. What do the position and velocity graphs look like? Slide 2-104
105 QuickCheck 2.20 A cart speeds up toward the origin. What do the position and velocity graphs look like? C. Slide 2-105
106 QuickCheck 2.21 A cart speeds up while moving away from the origin. What do the velocity and acceleration graphs look like? Slide 2-106
107 QuickCheck 2.21 A cart speeds up while moving away from the origin. What do the velocity and acceleration graphs look like? B. Slide 2-107
108 QuickCheck 2.22 Here is a motion diagram of a car speeding up on a straight road: The sign of the acceleration a x is A. Positive. B. Negative. C. Zero. Slide 2-108
109 QuickCheck 2.22 Here is a motion diagram of a car speeding up on a straight road: The sign of the acceleration a x is A. Positive. B. Negative. C. Zero. Speeding up means v x and a x have the same sign. Slide 2-109
110 QuickCheck 2.23 A cart slows down while moving away from the origin. What do the velocity and acceleration graphs look like? Slide 2-110
111 QuickCheck 2.23 A cart slows down while moving away from the origin. What do the velocity and acceleration graphs look like? C. Slide 2-111
112 QuickCheck 2.24 A cart speeds up while moving toward the origin. What do the velocity and acceleration graphs look like? Slide 2-112
113 QuickCheck 2.24 A cart speeds up while moving toward the origin. What do the velocity and acceleration graphs look like? C. Slide 2-113
114 QuickCheck 2.25 Which velocity-versus-time graph goes with this acceleration graph? Slide 2-114
115 QuickCheck 2.25 Which velocity-versus-time graph goes with this acceleration graph? E. Slide 2-115
116 The Moving Man The Creepy Car Slide 2-116
117 Section 2.5 Motion with Constant Acceleration
118 Reading Question 2.4 If an object is speeding up, A. Its acceleration is positive. B. Its acceleration is negative. C. Its acceleration can be positive or negative depending on the direction of motion. Slide 2-118
119 Reading Question 2.4 If an object is speeding up, A. Its acceleration is positive. B. Its acceleration is negative. C. Its acceleration can be positive or negative depending on the direction of motion. Slide 2-119
120 Motion with Constant Acceleration We can use the slope of the graph in the velocity graph to determine the acceleration of the rocket. Slide 2-120
121 Constant Acceleration Equations We can use the acceleration to find (v x ) f at a later time t f. We have expressed this equation for motion along the x-axis, but it is a general result that will apply to any axis. Slide 2-121
122 Constant Acceleration Equations The velocity-versus-time graph for constant-acceleration motion is a straight line with value (v x ) i at time t i and slope a x. The displacement Δx during a time interval Δt is the area under the velocity-versustime graph shown in the shaded area of the figure. Slide 2-122
123 Constant Acceleration Equations The shaded area can be subdivided into a rectangle and a triangle. Adding these areas gives Slide 2-123
124 Constant Acceleration Equations Combining Equation 2.11 with Equation 2.12 gives us a relationship between displacement and velocity: Δx in Equation 2.13 is the displacement (not the distance!). Slide 2-124
125 Constant Acceleration Equations For motion with constant acceleration: Velocity changes steadily: The position changes as the square of the time interval: We can also express the change in velocity in terms of distance, not time: Text: p. 43 Slide 2-125
126 Example 2.8 Coming to a stop As you drive in your car at 15 m/s (just a bit under 35 mph), you see a child s ball roll into the street ahead of you. You hit the brakes and stop as quickly as you can. In this case, you come to rest in 1.5 s. How far does your car travel as you brake to a stop? Slide 2-126
127 Example 2.8 Coming to a stop As you drive in your car at 15 m/s (just a bit under 35 mph), you see a child s ball roll into the street ahead of you. You hit the brakes and stop as quickly as you can. In this case, you come to rest in 1.5 s. How far does your car travel as you brake to a stop? PREPARE The problem statement gives us a description of motion in words. To help us visualize the situation, FIGURE 2.30 illustrates the key features of the motion with a motion diagram and a velocity graph. The graph is based on the car slowing from 15 m/s to 0 m/s in 1.5 s. Slide 2-127
128 Example 2.8 Coming to a stop (cont.) SOLVE We ve assumed that your car is moving to the right, so its initial velocity is (v x ) i = +15 m/s. After you come to rest, your final velocity is (v x ) f = 0 m/s. We use the definition of acceleration from Synthesis 2.1: An acceleration of 10 m/s 2 (really 10 m/s per second) means the car slows by 10 m/s every second. Now that we know the acceleration, we can compute the distance that the car moves as it comes to rest using the second constant acceleration equation in Synthesis 2.1: Slide 2-128
129 Example 2.8 Coming to a stop (cont.) ASSESS 11 m is a little over 35 feet. That s a reasonable distance for a quick stop while traveling at about 35 mph. The purpose of the Assess step is not to prove that your solution is correct but to use common sense to recognize answers that are clearly wrong. Had you made a calculation error and ended up with an answer of 1.1 m less than 4 feet a moment s reflection should indicate that this couldn t possibly be correct. Slide 2-129
130 Quadratic Relationships Text: p. 44 Slide 2-130
131 Example 2.9 Displacement of a drag racer A drag racer, starting from rest, travels 6.0 m in 1.0 s. Suppose the car continues this acceleration for an additional 4.0 s. How far from the starting line will the car be? Slide 2-131
132 Example 2.9 Displacement of a drag racer A drag racer, starting from rest, travels 6.0 m in 1.0 s. Suppose the car continues this acceleration for an additional 4.0 s. How far from the starting line will the car be? PREPARE We assume that the acceleration is constant, and the initial speed is zero, so the displacement will scale as the square of the time. SOLVE After 1.0 s, the car has traveled 6.0 m; after another 4.0 s, a total of 5.0 s will have elapsed. The initial elapsed time was 1.0 s, so the elapsed time increases by a factor of 5. The displacement thus increases by a factor of 5 2, or 25. The total displacement is Δx = 25(6.0 m) = 150 m ASSESS This is a big distance in a short time, but drag racing is a fast sport, so our answer makes sense. Slide 2-132
133 Section 2.6 Solving One-Dimensional Motion Problems
134 Problem-Solving Strategy Slide 2-134
135 Problem-Solving Strategy (cont.) Slide 2-135
136 Problem-Solving Strategy (cont.) Slide 2-136
137 Problem-Solving Strategy (cont.) Slide 2-137
138 The Pictorial Representation Text: p. 46 Slide 2-138
139 The Visual Overview The visual overview will consist of some or all of the following elements: A motion diagram. A good strategy for solving a motion problem is to start by drawing a motion diagram. A pictorial representation, as defined above. A graphical representation. For motion problems, it is often quite useful to include a graph of position and/or velocity. A list of values. This list should sum up all of the important values in the problem. Slide 2-139
140 Example 2.11 Kinematics of a rocket launch A Saturn V rocket is launched straight up with a constant acceleration of 18 m/s 2. After 150 s, how fast is the rocket moving and how far has it traveled? Slide 2-140
141 Example 2.11 Kinematics of a rocket launch A Saturn V rocket is launched straight up with a constant acceleration of 18 m/s 2. After 150 s, how fast is the rocket moving and how far has it traveled? PREPARE FIGURE 2.32 shows a visual overview of the rocket launch that includes a motion diagram, a pictorial representation, and a list of values. The visual overview shows the whole problem in a nutshell. The motion diagram illustrates the motion of the rocket. The pictorial representation (produced according to Tactics Box 2.2) shows axes, identifies the important points of the motion, and defines variables. Finally, we have included a list of values that gives the known and unknown quantities. In the visual overview we have taken the statement of the problem in words and made it much more precise. The overview contains everything you need to know about the problem. Slide 2-141
142 Example 2.11 Kinematics of a rocket launch (cont.) SOLVE Our first task is to find the final velocity. Our list of values includes the initial velocity, the acceleration, and the time interval, so we can use the first kinematic equation of Synthesis 2.1 to find the final velocity: Slide 2-142
143 Example 2.11 Kinematics of a rocket launch (cont.) SOLVE The distance traveled is found using the second equation in Synthesis 2.1: Slide 2-143
144 Problem-Solving Strategy for Motion with Constant Acceleration Text: p. 48 Slide 2-144
145 Example 2.12 Calculating the minimum length of a runway A fully loaded Boeing 747 with all engines at full thrust accelerates at 2.6 m/s 2. Its minimum takeoff speed is 70 m/s. How much time will the plane take to reach its takeoff speed? What minimum length of runway does the plane require for takeoff? Slide 2-145
146 Example 2.12 Calculating the minimum length of a runway A fully loaded Boeing 747 with all engines at full thrust accelerates at 2.6 m/s 2. Its minimum takeoff speed is 70 m/s. How much time will the plane take to reach its takeoff speed? What minimum length of runway does the plane require for takeoff? PREPARE The visual overview of FIGURE 2.33 summarizes the important details of the problem. We set x i and t i equal to zero at the starting point of the motion, when the plane is at rest and the acceleration begins. The final point of the motion is when the plane achieves the necessary takeoff speed of 70 m/s. The plane is accelerating to the right, so we will compute the time for the plane to reach a velocity of 70 m/s and the position of the plane at this time, giving us the minimum length of the runway. Slide 2-146
147 Example 2.12 Calculating the minimum length of a runway (cont.) SOLVE First we solve for the time required for the plane to reach takeoff speed. We can use the first equation in Synthesis 2.1 to compute this time: We keep an extra significant figure here because we will use this result in the next step of the calculation. Slide 2-147
148 Example 2.12 Calculating the minimum length of a runway (cont.) SOLVE Given the time that the plane takes to reach takeoff speed, we can compute the position of the plane when it reaches this speed using the second equation in Synthesis 2.1: Our final answers are thus that the plane will take 27 s to reach takeoff speed, with a minimum runway length of 940 m. Slide 2-148
149 Example 2.12 Calculating the minimum length of a runway (cont.) ASSESS Think about the last time you flew; 27 s seems like a reasonable time for a plane to accelerate on takeoff. Actual runway lengths at major airports are 3000 m or more, a few times greater than the minimum length, because they have to allow for emergency stops during an aborted takeoff. (If we had calculated a distance far greater than 3000 m, we would know we had done something wrong!) Slide 2-149
150 Section 2.7 Free Fall
151 Reading Question 2.5 A 1-pound ball and a 100-pound ball are dropped from a height of 10 feet at the same time. In the absence of air resistance A. The 1-pound ball wins the race. B. The 100-pound ball wins the race. C. The two balls end in a tie. D. There s not enough information to determine which ball wins the race. Slide 2-151
152 Reading Question 2.5 A 1-pound ball and a 100-pound ball are dropped from a height of 10 feet at the same time. In the absence of air resistance A. The 1-pound ball wins the race. B. The 100-pound ball wins the race. C. The two balls end in a tie. D. There s not enough information to determine which ball wins the race. Slide 2-152
153 Free Fall If an object moves under the influence of gravity only, and no other forces, we call the resulting motion free fall. Any two objects in free fall, regardless of their mass, have the same acceleration. On the earth, air resistance is a factor. For now we will restrict our attention to situations in which air resistance can be ignored. Apollo 15 lunar astronaut David Scott performed a classic experiment on the moon, simultaneously dropping a hammer and a feather from the same height. Both hit the ground at the exact same time something that would not happen in the atmosphere of the earth! Slide 2-153
154 Free Fall The figure shows the motion diagram for an object that was released from rest and falls freely. The diagram and the graph would be the same for all falling objects. Slide 2-154
155 Free Fall The free-fall acceleration always points down, no matter what direction an object is moving. Any object moving under the influence of gravity only, and no other force, is in free fall. Slide 2-155
156 Free Fall g, by definition, is always positive. There will never be a problem that uses a negative value for g. Even though a falling object speeds up, it has negative acceleration ( g). Because free fall is motion with constant acceleration, we can use the kinematic equations for constant acceleration with a y = g. g is not called gravity. g is the free-fall acceleration. g = 9.80 m/s 2 only on earth. Other planets have different values of g. We will sometimes compute acceleration in units of g. Slide 2-156
157 QuickCheck 2.26 A ball is tossed straight up in the air. At its very highest point, the ball s instantaneous acceleration a y is A. Positive. B. Negative. C. Zero. Slide 2-157
158 QuickCheck 2.26 A ball is tossed straight up in the air. At its very highest point, the ball s instantaneous acceleration a y is A. Positive. B. Negative. C. Zero. Slide 2-158
159 QuickCheck 2.27 An arrow is launched vertically upward. It moves straight up to a maximum height, then falls to the ground. The trajectory of the arrow is noted. At which point of the trajectory is the arrow s acceleration the greatest? The least? Ignore air resistance; the only force acting is gravity. Slide 2-159
160 QuickCheck 2.27 An arrow is launched vertically upward. It moves straight up to a maximum height, then falls to the ground. The trajectory of the arrow is noted. At which point of the trajectory is the arrow s acceleration the greatest? The least? Ignore air resistance; the only force acting is gravity. Same at all points. Slide 2-160
161 QuickCheck 2.28 An arrow is launched vertically upward. It moves straight up to a maximum height, then falls to the ground. The trajectory of the arrow is noted. Which graph best represents the vertical velocity of the arrow as a function of time? Ignore air resistance; the only force acting is gravity. Slide 2-161
162 QuickCheck 2.28 An arrow is launched vertically upward. It moves straight up to a maximum height, then falls to the ground. The trajectory of the arrow is noted. Which graph best represents the vertical velocity of the arrow as a function of time? Ignore air resistance; the only force acting is gravity. D Slide 2-162
163 Example 2.14 Analyzing a rock s fall A heavy rock is dropped from rest at the top of a cliff and falls 100 m before hitting the ground. How long does the rock take to fall to the ground, and what is its velocity when it hits? Slide 2-163
164 Example 2.14 Analyzing a rock s fall A heavy rock is dropped from rest at the top of a cliff and falls 100 m before hitting the ground. How long does the rock take to fall to the ground, and what is its velocity when it hits? PREPARE FIGURE 2.36 shows a visual overview with all necessary data. We have placed the origin at the ground, which makes y i = 100 m. Slide 2-164
165 Example 2.14 Analyzing a rock s fall (cont.) SOLVE Free fall is motion with the specific constant acceleration a y = g. The first question involves a relation between time and distance, a relation expressed by the second equation in Synthesis 2.1. Using (v y ) i = 0 m/s and t i = 0 s, we find We can now solve for t f : Now that we know the fall time, we can use the first kinematic equation to find (v y ) f : Slide 2-165
166 Example 2.14 Analyzing a rock s fall (cont.) ASSESS Are the answers reasonable? Well, 100 m is about 300 feet, which is about the height of a 30-floor building. How long does it take something to fall 30 floors? Four or five seconds seems pretty reasonable. How fast would it be going at the bottom? Using an approximate version of our conversion factor 1 m/s 2 mph, we find that 44.3 m/s 90 mph. That also seems like a pretty reasonable speed for something that has fallen 30 floors. Suppose we had made a mistake. If we misplaced a decimal point we could have calculated a speed of 443 m/s, or about 900 mph! This is clearly not reasonable. If we had misplaced the decimal point in the other direction, we would have calculated a speed of 4.3 m/s 9 mph. This is another unreasonable result, because this is slower than a typical bicycling speed. Slide 2-166
167 Summary: General Strategies Text: p. 55 Slide 2-167
168 Summary: General Strategies Text: p. 55 Slide 2-168
169 Summary: Important Concepts Text: p. 55 Slide 2-169
170 Summary: Important Concepts Text: p. 55 Slide 2-170
171 Summary: Important Concepts Text: p. 55 Slide 2-171
172 Summary: Applications Text: p. 55 Slide 2-172
173 Summary: Applications Text: p. 55 Slide 2-173
174 Summary: Applications Text: p. 55 Slide 2-174
175 Summary Text: p. 55 Slide 2-175
176 Summary Text: p. 55 Slide 2-176
177 Summary Text: p. 55 Slide 2-177
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