1. Chapter 1 Stop to Think / Quick Check Problems
This will be an excellent test of your understanding. I HIGHLY recommend you go through and UNDERSTAND each one of these.

2. Stop to Think
Car A
Car B
Motion diagrams are made of two cars. Both have the same time interval between photos. Which car, A or B, is going faster?
Answer: Car B
Reason: Each diagram has the same time interval between photos. As a result, the faster car would be the one that moves the most in each frame. In this case, car B is moving further in each frame and so it must be moving faster.
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3. Stop to Think
Three motion diagrams are shown. Which is a dust particle settling to the floor at a constant speed, which is a ball dropped from the roof of a building, and which is a descending rocket slowing to make a soft landing on mars?
Answer: A: Dropped ball. B: Dust particle. C: Rocket.
Reason: The dropped ball will accelerate due to the presence of gravity and increase in speed. As a result, the motion diagram shows an object that increases in distance each frame as it moves toward the floor as in A. The dust particle settles to the floor at a constant speed. As a result. The frames in a motion diagram will be equally spaced as they are in B. The rocket leaves the ground, starting at rest and accelerating due to the rocket engine. We will see the frames increasing in separation as height increases as is shown in C.

4. Stop to Think
You start at a positive position along the x-axis and undergo a negative displacement. Your final position
Is positive
Is negative
Could be either positive or negative
Answer: C.
Reason: It depends upon where you start. Say you start at x=+3. You could undergo a negative displacement of -2, giving a final position of x = +1. You could also undergo a negative displacement of -5, giving a final position of x = -2.

5. Stop to Think
Maria is at position x = 23 m. She then undergoes a displacement x = –50 m. What is her final position?
–27 m
–50 m
23 m
73 m
Answer: A
Reason: Δx = x_f – x_i. Thus, x_f = Δx + x_i = -50m + 23m = -27m.
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6. Chapter 2 Quick Check Problems

7. Quick Check 2.2
Here is a motion diagram of a car moving along a straight road:
Which velocity-versus-time graph matches this motion diagram?
None of the above.
Answer: C
Reason: The object is moving to the right (positive direction). For several frames it is moving at some constant speed. Then, for the next frames, it is suddenly moving at some faster speed.
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8. Quick Check 2.3
Here is a motion diagram of a car moving along a straight road:
Which velocity-versus-time graph matches this motion diagram?
Answer: D
Reason: Same as before but negative. The object moves with some small negative speed and then suddenly with a faster (larger) negative speed.
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9. Quick Check 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?
Answer: C
Reason: The first segment shows the position is not changing. If the position isn’t changing, there is no velocity, so the velocity graph will be zero for this segment. For the second segment, the slope is negative and constant (a straight line). Therefore, the velocity will be negative and constant.
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10. Quick Check 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?
Answer: D
Reason: The first segment shows a constant positive velocity. This implies that the position will change at a constant rate in the positive direction. The second segment has zero velocity. As a result, the object is not moving and its position will not change from where it was at the end of the first segment. The result is then answer D.
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11. Quick Check 2.7
Which velocity-versus-time graph goes with this position graph?
Answer: C
Reason: Break this motion into three segments. The first segment, from the beginning to where it crosses the time axis, has a positive, constant slope. As a result, the velocity is positive and constant. The second segment still has a positive slope, but its slope is steadily decreasing. As a result, the velocity is still positive, but it is decreasing at a constant rate. The third segment is the flat portion at the end. Here, the slope is zero and, thus, the velocity is zero. Hence, the answer must be C.
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12. Quick Check 2.8 Here is a position graph of an object:
At t = 1.5 s, the object’s velocity is
40 m/s
20 m/s
10 m/s
–10 m/s
None of the above
Answer: B
Reason: Find the slope of the line at t = 1.5s. Slope = rise/run = (20m-0m)/(2s-1s) = 20m/1s = 20m/s.
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13. Stop to Think 2.2
Four objects move with the velocity vs. time graphs shown. Which object has the largest displacement between t = 0 s and t = 2 s?
Answer: B
Reason: Find the area under the curve. Whichever has the largest area has the largest displacement. Area from A: (1m/s)*(2s) = 2m. Area from B: (2m/s)*(2s) = 4m. Area from C: (2m/s)*(1s) = 2m. Area from D: (1.5m/s)*(1.5s) = 2.25m. Thus, B has the greatest area/displacement.

14. Quick Check 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
20 m
16 m
12 m
8 m
4 m
Answer: C
Reason: Find the area under the curve. To do this, break it up into a rectangle from 0s-2s and a triangle from 2s-4s. Area of rectangle = (4m/s)*(2s) = 8m. Area of triangle = 0.5*(2s)*(4m/s) = 4m. Total areal = 8m + 4m = 12m.
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15. Quick Check 2.9 When do objects 1 and 2 have the same velocity?
At some instant before time t0
At time t0
At some instant after time t0
Both A and B
Never
Answer: A
Reason: Velocity is equal to the slope of a position graph. If it asks for the same velocity, its asking for the same slope. In this graph, the slopes are the same well before t_0. Note: the POSITIONS are the same at t_0.
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16. Quick Check 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.
Answer: Zero displacement: D. Zero speed: B & E. Increasing speed: C. Decreasing speed: A.
Reason: In a position graph, displacement is zero when the object is back at its origin. In this case, that is when x = 0 (along the time axis). This happens only at point D. The speed is zero when the slope is zero. The slope is zero at B and E. The speed is increasing if the slope (steepness of the line) is increasing. The slope is only getting steeper at C. The speed is decreasing wherever the slope is decreasing. This occurs only at point A where the slope is going from some positive value down to zero at B.
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17. Quick Check 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?
Answer: B
Reason: The object moves with some velocity represented by the green vector. It has some magnitude (length). The small acceleration points opposite to its motion so it will be slowing down (decelerating). As a result, the vector arrow that comes next will be slightly smaller, but still pointing down. The correct answer is B.
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18. Quick Check 2.14
The motion diagram shows a particle that is slowing down. The sign of the position x and the sign of the velocity vx are:
Position is positive, velocity is positive.
Position is positive, velocity is negative.
Position is negative, velocity is positive.
Position is negative, velocity is negative.
Answer: B
Reason: The object is always right of the origin, thus, its position is always positive. Its velocity, however, is negative since it points to the left. The acceleration arrows are not needed to solve this one.
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19. Example Problem
A ball moving to the right traverses the ramp shown below. Sketch a graph of the velocity versus time, and, directly below it, using the same scale for the time axis, sketch a graph of the acceleration versus time.
Answer: The velocity graph starts with a constant value of v with time with a positive value of v. Then when the ball reaches the top of the ramp (label as point A) the velocity increases linearly until the ball reaches the bottom of the ramp (label as point B) at which point the velocity remains constant at the new high value. The acceleration graph starts at zero and continues at zero until point A at which point it jumps up to a constant value. With tilted axes, the x-direction points along the slope and so it will be a positive acceleration. The acceleration remains constant until point B at which point it returns to zero.
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20. Quick Check 2.15
The motion diagram shows a particle that is slowing down. The sign of the acceleration ax is:
Acceleration is positive.
Acceleration is negative.
Answer: A
Reason: It points in the positive direction (to the right).
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21. Quick Check 2.16
A cyclist riding at 20 mph sees a stop sign and 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?
Answer: B
Reason: As the person slows down to a stop I 4 seconds, there should be 4 dots, getting closer and closer together as it moves to the right. C is ruled out already as it shows 6 dots for this phase of motion. Since it is moving right and slowing down, the acceleration will point to the left (opposite to its motion). This rules out option A and D. Thus we already know B is the correct answer. After the 4 seconds of slowing down, he then speeds up for 6 seconds. The dots should get further apart and the acceleration will point in the direction of his motion (to the right) since he is speeding up. This fits B perfectly, confirming our answer.
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22. Quick Check 2.17
These four motion diagrams show the motion of a particle along the x-axis.
Which motion diagrams correspond to a positive acceleration?
Which motion diagrams correspond to a negative acceleration?
Answer: Positive acceleration: A & C. Negative acceleration: B & D.
Reason: For A, the object moves in the positive direction and speeds up. Since it is speeding up, acceleration points along its motion and so it also points in the positive (to the right) direction. For B, the object is moving in the positive direction, but it is slowing down. The acceleration then points opposite to its motion to the left (negative direction). For C, the object is moving in the negative direction and slowing down. Thus, the acceleration points opposite to its motion and points right (positive direction). For D, the object is moving to the left and is speeding up. Since it is speeding up, the acceleration points along its motion to the left (negative direction).
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23. Quick Check 2.21
A cart speeds up while moving away from the origin. What do the velocity and acceleration graphs look like?
Answer: B
Reason: The cart moves in the positive direction so its velocity will be positive. The speed is increasing so it will move away from the time axis where the speed is zero. Also, since its speed is increasing, the acceleration will point in the same direction as its motion in the positive direction. The results are those of option B.
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24. Quick Check 2.23
A cart slows down while moving away from the origin. What do the velocity and acceleration graphs look like?
Answer: C
Reason: The cart moves in the positive direction so its velocity will be positive. The speed, however, is decreasing so it will have move toward the time axis where the speed is zero. Also, since its speed is decreasing, the acceleration will point in the opposite direction of its motion (in the negative direction). The results are those of option C.
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25. Quick Check 2.24
A cart speeds up while moving toward the origin. What do the velocity and acceleration graphs look like?
Answer: C
Reason: The cart moves in the negative direction so its velocity will be negative. The speed is increasing so it will move away from the time axis where the speed is zero. Also, since its speed is increasing, the acceleration will point in the same direction as its motion in the negative direction. The results are those of option C.
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26. Quick Check 2.25
Which velocity-versus-time graph goes with this acceleration graph?
Answer: E
Reason: The first segment is negative and constant. As a result, the velocity will have a negative slope and will change at a constant rate ( a straight line). The second segment instantly changes to a positive constant acceleration. Here, the slope of the velocity graph would be positive and constant. Each segment lasts the same amount of time and has the same magnitude so the velocity graph will end at the same velocity that it started with (zero in this case). Note: for this problem, the velocity graph lies entirely below the horizontal axis simple because it is assumed it starts at zero at time zero.
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27. Quick Check 2.26
A ball is tossed straight up in the air. At its very highest point, the ball’s instantaneous acceleration
ay is
Positive.
Negative.
Zero.
Answer: B
Reason: The vertical acceleration of any object only under the influence of gravity is -9.8 m/s^2 throughout its entire free fall motion. Even though it has no velocity at its peak, the velocity CHANGES DIRECTION, and so there is still an acceleration present. The acceleration due to gravity doesn’t just suddenly vanish when the object is at its max height.
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28. Quick Check 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.
Answer: Trick question. Same at all points.
Reason: Again, the acceleration of any object in free fall is -9.8 m/s^2., regardless of where in its path the object lies.
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29. Quick Check 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.
Answer: D
Reason: First, the arrow moves upward in the positive direction. As a result, the velocity will be positive. The velocity is, however, decreasing at a constant rate (by 9.8 m/s every second) as it goes up, approaching zero at its peak. So, its velocity graph will approach zero velocity at a constant rate. Then, the arrow begins its descent in the negative direction (so it will have a negative velocity now). The arrow is speeding up from zero to some greater (more negative) value, at the same constant rate as before. The total result is that of D.
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30. Chapter 3 Quick Check Problems

31. Quick Check 3.1 Given vectors and , what is ? Answer: A
Reason: See diagram in slide. Use tip-to-tail method and draw the resultant vector from the initial to the final point.
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32. Quick Check 3.3 Given vectors and , what is ? Answer: D
Reason: See diagram in slide. Use tip-to-tail method and draw the resultant vector from the initial to the final point. Recall that vector subtraction is the same as vector addition but with the negative vector having opposite direction.
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33. Quick Check 3.4 Which of the vectors in the second row shows 2  ?
Answer: A
Reason: See diagram in slide. Use tip-to-tail method and draw the resultant vector from the initial to the final point. 2A is simply twice as long as A. Recall that vector subtraction is the same as vector addition but with the negative vector having opposite direction.
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34. Quick Check 3.5
A particle undergoes acceleration while moving from point 1 to point 2. Which of the choices shows the velocity vector as the object moves away from point 2? A.
Answer: C
Reason: See diagram in slide. Recall that a = Δ v / Δ t = (v_f – v_i) / Δt or a = (v_2 – v_1) / Δt. Acceleration points in the same direction as Δv (given by my blue arrows). In C, the Δ v points down, matching the given diagram where a points down. B. C.
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35. Quick Check 3.6
The diagram shows three points of a motion diagram. The particle changes direction with no change of speed. What is the acceleration at point 2?
Answer: B
Reason: See diagram in slide. Again, acceleration will point in the direction of Δ v. Look over the tactics box 3.2 on page 69 of your text. Lets call the vector connecting points 1 and 2 vector v_1 and the other v_2. We simple find what v_2 – v_1 looks like and draw a in the direction of Δ v . Note that the answer is not C because the the lengths of the vectors are the same…but all of the vector v_1 lies in the x-direction while only part of vector v_2 lies in the x-direction. As a result, v_1 will extend a little bit past the start of v_2 and we get the resultant as shown.
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36. Quick Check 3.7 What are the x- and y-components of this vector? 3, 2
2, 3
–3, 2
2, –3
–3, –2
Answer: B
Reason: The vector points two units to the right and three units up. Hence, the components are 2, 3.
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37. Quick Check 3.8 What are the x- and y-components of this vector? 3, 4
4, 3
–3, 4
4, –3
3, –4
Answer: E
Reason: The vector points three units to the right and four units down. Hence, the components are 3,-4.
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38. Quick Check 3.9 What are the x- and y-components of vector C? 1, –3
–3, 1
1, –1
–4, 2
2, –4
Answer: D
Reason: The vector points four units to the left and two units up. Hence, the components are -4,2. Note: it seems more confusing because it is crossing axes, but that does not influence the components of a vector.
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39. Quick Check 3.10 The angle Φ that specifies the direction of vector is
tan–1(Cx/Cy)
tan–1 (Cy/Cx)
tan–1 (|Cx|/Cy)
tan–1 (|Cx|/|Cy|)
tan–1 (|Cy|/|Cx|)
Answer: D
Reason: Both the x and the y components are negative, so both need to be absolute value’d when finding the angle phi. Tangent in this case would equal C_x / c_y. The result is D.
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40. Quick Check 3.11
The following vector has length 4.0 units. What are the x- and y-components?
3.5, 2.0
–2.0, 3.5
–3.5, 2.0
2.0, –3.5
–3.5, –2.0
Answer: B
Reason: You are welcome to use your sin and cos trig functions to calculate the answer, but you need not to. You may answer this without any calculations. First note that the x-component is negative and the y-component is positive. This rules out options A, D, and E. Now it is a matter of determining which component will be greater. In this case, the y-component is larger and so the correct answer is B.
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41. Quick Check 3.14
A ball rolls up the ramp, then back down. Which is the correct acceleration graph?
Answer: D
Reason: Use tilted axes so that the x-axis points up the ramp. As the ball moves up the ramp, it is slowing down. As a result, the acceleration points opposite to its direction of motion (down the ramp in the negative direction) and at a constant rate. This gives a straight line with zero slope in the negative region of the acceleration graph. Now, on the way down the ramp, the ball will be moving in the negative direction and it will speed up. Since it speeds up, the acceleration points in the direction of the motion (in the negative direction) and still at the same constant rate. The total result is a straight line with zero slope in the negative region of the acceleration graph. Thus the answer is D. It is not E since the acceleration doesn’t suddenly disappear at the top of its motion (remember there is an acceleration if the velocity changes direction).
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42. Quick Check 3.15
A factory conveyor belt rolls at 3 m/s. A mouse sees a piece of cheese directly across the belt and heads straight for the cheese at 4 m/s. What is the mouse’s speed relative to the factory floor?
1 m/s
2 m/s
3 m/s
4 m/s
5 m/s
Answer: E
Reason: The mouse moves to the right at 4 m/s and down at 3 m/s. His total motion, therefore, relative to someone watching from afar, would be what is shown by the blue vector arrow in the diagram in the slide. Pythagorean theorem may be used to determine the total motion. Recall: A^2 + B^2 = C^2 where C is the hypotenuse of a right triangle and A and B are the sides. Here we get: 3^2 + 4^2 = = 25. Therefore, C = 5 m/s.
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43. Quick Check 3.17
A 100-g ball rolls off a table and lands 2.0 m from the base of the table. A 200-g ball rolls off the same table with the same speed. It lands at distance
1.0 m
Between 1 m and 2 m
2.0 m
Between 2 m and 4 m
4.0 m
Answer: C
Reason: Recall that objects in projectile motion all fall with a trajectory that follows the same type of path. Neglecting air resistence, all objects fall at the same rate regardless of mass (remember the feather / bowling ball video?). Furthermore, think about all the equations we discussed dealing with the motion of objects. None of them have mass terms in them, indicating that mass is negligible. The answer is then the same for both objects and we arrive at option C.
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