Module 3 Accelerated Motion Lesson 1 Acceleration Practice Problems PDF

Module 3 continued MODULE 3 Accelerated Motion...

Module 3 continued MODULE 3 Accelerated Motion c. 0.0 s to 40.0 s Lesson 1 Acceleration v2  v1 a  Practice Problems t2  t1 p. 63 0.0 m/s  0.0 m/s  40.0 s  0.0 s 1. The velocity-time graph in Figure 8 describes Steven’s motion as he walks  0.0 m/s2 along the midway at the state fair. Sketch 4. CHALLENGE Plot a v-t graph the corresponding motion diagram. Include representing the following motion: An velocity vectors in your diagram. elevator starts at rest from the ground floor of a three-story shopping mall. It accelerates upward for 2.0 s at a rate of 0.5 m/s2, continues up at a constant velocity of 1.0 m/s for 12.0 s, and then 2. Use the v-t graph of the toy train in slows down with a constant downward Figure 9 to answer these questions. acceleration of 0.25 m/s2 for 4.0 s as it reaches the third floor. a. When is the train’s speed constant? 5.0 to 15.0 s b. During which time interval is the train’s acceleration positive? 0.0 to 5.0 s c. When is the train’s acceleration most negative? 15.0 to 20.0 s Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3. Refer to Figure 9 to find the average acceleration of the train during the following time intervals. Practice Problems p. 64 a. 0.0 s to 5.0 s v2  v1 5. A race car’s forward velocity increases a  t2  t1 from 4.0 m/s to 36 m/s over a 4.0-s time interval. What is its average acceleration? 10.0 m/s  0.0 m/s  5.0 s  0.0 s v 36 m/s  4.0 m/s a    8.0 m/s2  2.0 m/s 2 t 4.0 s  8.0 m/s2 forward b. 15.0 s to 20.0 s v2  v1 6. The race car in the previous problem slows a  t2  t1 from 36 m/s to 15 m/s over 3.0 s. What is 4.0 m/s  10.0 m/s its average acceleration?  20.0 s  15.0 s v 15 m/s  36 m/s a     7.0 m/s2  1.2 m/s2 t 3.0 s  7.0 m/s2 backward Inspire Physics 1 Solutions Manual Module 3 continued 7. A bus is moving west at 25 m/s when the Lesson 1 Check Your Progress driver steps on the brakes and brings the p. 65 bus to a stop in 3.0 s. a. What is the average acceleration of the 11. Describing Motion What are three ways bus while braking? an object can accelerate? v 0.0 m/s  25 m/s speed up, slow down, change a   t 3.0 s direction   8.3 m/s2  8.3 m/s2 east 12. Position-Time and Velocity-Time b. If the bus took twice as long to stop, Graphs Two joggers run at a constant how would the acceleration compare velocity of 7.5 m/s east. Figure 10 shows with what you found in part a? the positions of both joggers at time t  0. half as great (4.2 m/s2 east) a. What would be the difference(s) in the position-time graphs of their motion? 8. A car is coasting backward downhill at a speed of 3.0 m/s when the driver gets the Both lines would have the same engine started. After 2.5 s, the car is slope, but they would have moving uphill at 4.5 m/s. If uphill is chosen different y-intercepts, 15 m and as the positive direction, what is the car’s 15 m. average acceleration? b. What would be the difference(s) in their v 4.5 m/s  (3.0 m/s) velocity-time graphs? a   t 2.5 s Their velocity-time graphs would be identical.  3.0 m/s2 13. Velocity-Time Graph Sketch a velocity- time graph for a car that goes east at 9. Rohith has been jogging east toward the 25 m/s for 100 s, then west at 25 m/s for bus stop at 3.5 m/s when he looks at his another 100 s. watch and sees that he has plenty of time before the bus arrives. Over the next 10.0 s, he slows his pace to a leisurely 0.75 m/s. What was his average Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. acceleration during this 10.0 s? v 0.75 m/s  3.5 m/s a   t 10.0s   0.28 m/s  0.28 m/s2 west 2 10. CHALLENGE If the rate of continental drift were to abruptly slow from 1.0 cm/y to 0.5 cm/y over the time interval of a year, what would be the average acceleration? v 0.5 cm/yr  1.0 cm/yr a   t 1.0 yr   0.5 cm/yr 2  0.5 cm/yr 2 in the direction opposite the drift Solutions Manual 2 Inspire Physics  Module 3 continued 14. Average Velocity and Average 15. Critical Thinking A police officer clocked Acceleration A canoeist paddles a driver going 32 km/h over the speed limit upstream at a velocity of 2.0 m/s for 4.0 s just as the driver passed a slower car. and then floats downstream at 4.0 m/s for When the officer stopped the car, the driver 4.0 s. argued that the other driver should get a ticket as well. The driver said that the cars a. What is the average velocity of the must have been going the same speed canoe during the 8.0-s time interval? because they were observed next to each Choose a coordinate system with other. Is the driver correct? Explain with a the positive direction upstream. sketch and a motion diagram. vi  vf v  No; the cars had the same position, 2 not velocity. To have the same 2.0 m/s  (4.0 m/s) velocity, they would have had to have  2 the same relative position for a length  1.0 m/s of time.  1.0 m/s downstream b. What is the average acceleration of the canoe during the 8.0-s time interval? v a  t vf  vi  t (4.0 m/s)  (2.0 m/s)  8.0 s  0.75 m/s2  0.75 m/s2 downstream Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Inspire Physics 3 Solutions Manual Module 3 continued 18. If a car accelerates from rest at a constant Lesson 2 Motion with rate of 5.5 m/s2 north, how long will it take Constant Acceleration for the car to reach a velocity of 28 m/s north? Practice Problems p. 68 vf vi  at vf  vi so t  16. A golf ball rolls up a hill toward a miniature- a golf hole. Assume the direction toward the 28 m/s north  0.0 m/s hole is positive.  5.5 m/s2 north a. If the golf ball starts with a speed of  5.1 s 2.0 m/s and slows at a constant rate of 0.50 m/s2, what is its velocity after 19. CHALLENGE A car slows from 22 m/s to 2.0 s? 3.0 m/s at a constant rate of 2.1 m/s2. How vf  vi  at many seconds are required before the car is traveling at a forward velocity of 3.0 m/s?  2.0 m/s  (0.50 m/s2 )(2.0 s) Let the forward direction be positive.  1.0 m/s vf  vi  at vf  vi b. What is the golf ball’s velocity if the so t  constant acceleration continues for a 6.0 s? 3.0 m/s  22 m/s  vf  vi  at 2.1 m/s2  9.0 s  2.0 m/s  (0.50 m/s2 )(6.0 s)  1.0 m/s Practice Problems c. Describe the motion of the golf ball in p. 69 words and with a motion diagram. The ball’s velocity decreased in the 20. The graph in Figure 13 describes the first case. In the second, the ball motion of two bicyclists, Akiko and Brian, slowed to a stop and then began who start from rest and travel north, increasing their speed with a constant Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. rolling back down the hill. acceleration. What was the total displacement of each bicyclist during the time shown for each? Hint: Use the area of a triangle: area  (1/2)(base)(height). xA  9.0 m north; xB  8.0 m north 17. A bus traveling at 30.0 km/h east has a constant increase in speed of 1.5 m/s2. 21. The motion of two people, Carlos and What velocity does it reach 6.8 s later? Diana, moving south along a straight path is described by the graph in Figure 14. vf  vi  at What is the total displacement of each  30.0 km/h person during the first 4.0-s interval shown  1km  3600 s  on the graph?  (1.5 m/s2 )(6.8 s)     1000 m  1h  xC  8.0 m south; xD  5.0 m south  67 km/h east Solutions Manual 4 Inspire Physics  Module 3 continued 22. CHALLENGE A car, just pulling onto a Practice Problems straight stretch of highway, has a constant p. 71 acceleration from 0 m/s to 25 m/s west in 12 s. 23. A skateboarder is moving at a constant a. Draw a v-t graph of the car’s motion. speed of 1.75 m/s when she starts up an incline that causes her to slow down with a constant acceleration of 0.20 m/s2. How much time passes from when she begins to slow down until she begins to move back down the incline? vf vi  at vf  vi 0.0 m/s  1.75 m/s t   8.8 s a  0.20 m/s2 24. A race car travels on a straight racetrack with a forward velocity of 44 m/s and slows at a constant rate to a velocity of 22 m/s b. Use the graph to determine the car’s over 11 s. How far does it move during this displacement during the 12.0-s time time? interval. The displacement is the area under Let the positive direction be forward. the velocity-time graph. vf  vi  at 1 vf  vi x  (base)(height) a 2 t 1 22 m/s  44 m/s  (12.0 s)(25 m/s)  2 11 s  150 m west   2.0 m/s 1 c. Another car is traveling along the same xf  xi  v itf  atf 2 stretch of highway. It travels the same 2  0  (44 m/s)(11 s)  Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. distance in the same time as the first car, but its velocity is constant. Draw a 1 ( 2.0 m/s2 )(11 s)2 v-t graph for this car’s motion. 2  360 m 25. A car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achieve the final speed? v f2  v i2 a  2 x (25 m/s)2  (15 m/s)2   1.6 m/s2 (2)(125 m) d. Explain how you knew this car’s t  vf a vi velocity. 25 m/s  15 m/s The displacement was the same for   6.3 s 1.6 m/s2 both cars. For the second car, then, v  x / t  150 m  12 s  13 m west (rounding to the correct number of significant figures). Inspire Physics 5 Solutions Manual Module 3 continued 26. A bike rider pedals with constant 29. A man runs along the path shown in acceleration to reach a velocity of Figure 17. From point A to point B, he 7.5 m/s north over a time of 4.5 s. During runs at a forward velocity of 4.5 m/s for the period of acceleration, the bike’s 15.0 min. From point B to point C, he runs displacement is 19 m north. What was the up a hill. He slows down at a constant rate initial velocity of the bike? of 0.050 m/s2 for 90.0 s and comes to a stop at point C. What was the total distance vi  vf v  the man ran? 2 (v i  v f )t x AB  v ABt AB x  v t  2  (4.5 m/s)(15.0 min)(60 s/min) 2 x  4050 m (carrying extra digits) so v i   vf t 1 2(19 m) x BC  v it f  at f2   7.5 m/s 2 4.5 s  (4.5 m/s)(90.0 s)  0.94 m/s north 1  (0.050 m/s2 )(90.0 s)2 2 27. Challenge The car in Figure 16  203 m (carrying an extra digit) travels west with a forward acceleration of 0.22 m/s2. What was the car’s velocity (vi) x AB  xBC  4050 m  203 m at point xi if it travels a distance of 350 m  4.3  102 m in 18.4 s? 30. You start your bicycle ride at the top of a 1 xf  xi  v i t f  at f 2 hill. You coast down the hill at a constant 2 acceleration of 2.00 m/s2. When you get to 1 2 ( x f  x )  at f the bottom of the hill, you are moving at vi  2 18.0 m/s, and you pedal to maintain that tf speed. If you continue at this speed for 1 1.00 min, how far will you have gone from 350 m (0.22 m/s2 )(18.4 s)2  2 the time you left the hilltop? 18.4 s Part 1: Constant acceleration:  17 m/s west Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. v f 2  v i 2  2a ( x f  x i ) but x i  0 m and v i  0 m/s Practice Problems vf2 (18.0 m/s)2 p. 73 x1  xf   2a (2)(2.00 m/s2 ) 28. A car with an initial velocity of 24.5 m/s  81.0 m east has an acceleration of 4.2 m/s2 west. Part 2: Constant velocity: What is its displacement at the moment x 2  vt  (18.0 m/s)(60.0 s) that its velocity is 18.3 m/s east?  1.08  103 m Let east be the positive direction. x  x1  x 2 v 2  v i2  81.0 m  1.08  103 m x  f 2a  1.16  103 m (18.3 m/s)2  (24.5 m/s)2  (2)(4.2 m/s2 )  32 m east Solutions Manual 6 Inspire Physics  Module 3 continued 31. Sunee is training for a 5.0-km race. She Lesson 2 Check Your Progress starts out her training run by moving at a p. 73 constant pace of 4.3 m/s for 19 min. Then she accelerates at a constant rate until she 33. Displacement Given initial and final crosses the finish line 19.4 s later. What is velocities and the constant acceleration of her acceleration during the last portion of an object, what mathematical relationship the training run? would you use to find the displacement? Let Sunee’s initial direction of motion vf2  v i 2  2a  x be positive. Part 1: Constant velocity: x  vt 34. Acceleration A woman driving west along a straight road at a speed of 23 m/s sees a  (4.3 m/s)(19 min)(60 s/min) deer on the road ahead. She applies the  4902 m in the positive direction brakes when she is 210 m from the deer. If Part 2: Constant acceleration: the deer does not move and the car stops 1 right before it hits the deer, what is the x f  x i  v it  at 2 2 acceleration provided by the car’s brakes? 2( x f  x i  v it ) Let the positive direction be west. a t2 v 2  vi 2 (0.0 m/s)2  (23 m/s)2 (2)(5.0  103 m  4902 m  a f  2( x f  x i ) (2)(210 m) (4.3 m/s)(19.4 s))   1.3 m/s2  1.3 m/s2 east (19.4 s)2  0.077 m/s2 in the positive direction 35. Distance The airplane in Figure 18 starts from rest and accelerates east at a constant 3.00 m/s2 for 30.0 s before leaving 32. CHALLENGE Sekazi is learning to ride a the ground. bike without training wheels. His father a. What was the plane’s displacement pushes him with a constant acceleration of (x)? 0.50 m/s2 east for 6.0 s. Sekazi then travels at 3.0 m/s east for another 6.0 s before Let the positive direction be east. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. falling. What is Sekazi’s displacement? 1 x  v it  at 2 Solve this problem by constructing a 2 velocity-time graph for Sekazi’s motion and  (0.0 m/s)(30.0 s) computing the area underneath the graphed line. 1  (3.00 m/s2 )(30.0 s)2 2  1.35  103 m east b. How fast was the airplane going when it took off? vf  vi  atf  0.0 m/s  (3.00 m/s2)(30.0 s) Let the positive direction be east.  90.0 m/s Part 1: Constant acceleration: 1 x 1  (3.0 m/s)(6.0 s)  9.0 m east 2 Part 2: Constant velocity: x 2  (3.0 m/s)(12.0 s  6.0 s)  18 m east x  x1  x2  9.0 m east  18 m east  27 m east Inspire Physics 7 Solutions Manual Module 3 continued 36. Distance An in-line skater accelerates 39. Graphs A sprinter walks to the starting from 0.0 m/s to 5.0 m/s in 4.5 s, then blocks at a constant speed, then waits. continues at this constant speed for When the starting pistol sounds, she another 4.5 s. What is the total distance accelerates rapidly until she attains a traveled by the in-line skater? constant velocity. She maintains this velocity until she crosses the finish line, Accelerating: and then she slows to a walk, taking more  v  vf t x f  vt f   i time to slow down than she did to speed up  f  2  at the beginning of the race. Sketch a 0.0 m/s  5.0 m/s    velocity-time and a position-time graph to  (4.5 s) represent her motion. Draw them one  2   11.25 m above the other using the same time scale. Indicate on your position-time graph where Constant speed: the starting blocks and finish line are. xf  v ftf  (5.0 m/s)(4.5 s)  22.5 m total distance  11.25 m  22.5 m  34 m 37. Final Velocity A plane travels 5.0×102 m north while being accelerated uniformly from rest at the rate of 5.0 m/s2. What final velocity does it attain? vf2  vi2  2a(xf  xi) but vi  0 m/s and xi  0 m, so Lesson 3 Free Fall vf2  2axf Practice Problems vf  2(5.0 m/s2 )(5.0  102 m) p. 78  71 m/s north 40. A construction worker accidentally drops a brick from a high scaffold. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 38. Final Velocity An airplane accelerated uniformly from rest at the rate of 5.0 m/s2 a. What is the velocity of the brick after south for 14 s. What final velocity did it 4.0 s? attain? Let upward be the positive vf  vi  atf  0 m/s  (5.0 m/s2)(14 s) direction.  7.0101 m/s south vf  vi  at, a  9.8 m/s2 vf  0.0 m/s  (9.8 m/s2)(4.0 s)  39 m/s  39 m/s downward b. How far does the brick fall during this time? 1 x  vit  2 at 2  1    0   2  (9.8 m/s2)(4.0 s)2  78 m The brick falls 78 m. Solutions Manual 8 Inspire Physics  Module 3 continued 41. Suppose for the previous problem you 44. You decide to flip a coin to determine choose your coordinate system so that the whether to do your physics or English opposite direction is positive. homework first. The coin is flipped straight up. a. What is the brick’s velocity after 4.0 s? Now the positive direction is a. What are the velocity and acceleration downward. of the coin at the top of its trajectory? vf  vi  at, a  9.8 m/s2 vtop  0 m/s; vf  0.0 m/s  (9.8 m/s2)(4.0 s) atop  9.8 m/s downward  39 m/s  39 m/s upward b. If the coin reaches a high point of b. How far does the brick fall during this 0.25 m above where you released it, time? what was its initial speed? 1 vf2  vi2  2ax x  vit  2 at2, a  9.8 m/s2 vi  vf  2g x where a  g 2  (0.0 m/s)(4.0 s) and vf  0 at the height of the toss.  1   vi  2g x   2  (9.8 m/s2)(4.0 s)2 2  78 m  (2)(9.8 m/s )(0.25) m The brick still falls 78 m.  2.2 m/s 42. A student drops a ball from a window 3.5 m c. If you catch it at the same height as you above the sidewalk. How fast is it moving released it, how much time was it in the when it hits the sidewalk? air? vf2  vi2  2ax, a  g and vi  0 vf  vi  at 2gx v f  vi so vf  2 t  g  (2)(9.8 m/s )(3.5 m) 2.2 m/s  2.2 m/s  8.3 m/s 9.8 m/s2  43. A tennis ball is thrown straight up with an  0.45 s Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. initial speed of 22.5 m/s. It is caught at the 45. CHALLENGE A basketball player is same distance above the ground. holding a ball in her hands at a height of a. How high does the ball rise? 1.5 m above the ground. She drops the ball, and it bounces several times. After the a  g first bounce, the ball only returns to a at the maximum height, vf  0 height of 0.75 m. After the second bounce, vf2  vi2  2ax becomes the ball only returns to a height of 0.25 m. vi2  2gx a. Suppose downward is the positive v i2 (22.5 m/s)2  2  26 m direction. What would the shape of a x  2g (2)(9.8 m/s ) velocity-time graph look like for the first b. How long does the ball remain in the two bounces? air? Hint: The time it takes the ball to The velocity-time graph would be rise equals the time it takes to fall. straight line segments that start at Rise time: the origin and then rise, fall, and rise again. vf  vi  at but a  g and vf  0, so b. What would be the shape of a position- v 22.5 m/s time graph for the first two bounces? t i   2.3 s g 9.8 m/s2 The graph would start at the origin The fall time equals the rise time, and have an inverted parabolic so the time to remain in the air is shape. tair  2trise  (2)(2.3 s)  4.6 s Inspire Physics 9 Solutions Manual Module 3 continued Lesson 3 Check Your Progress 49. Maximum Height and Flight Time The free-fall acceleration on Mars is about one- p. 79 third that on Earth. Suppose you throw a ball upward with the same velocity on Mars 46. Free Fall Suppose you hold a book in one as on Earth. hand and a flat sheet of paper in your other hand. You drop them both, and they fall to a. How would the ball’s maximum height the ground. Explain why the falling book is compare to that on Earth? a good example of free fall, but the paper is Let M  Mars and E  Earth. not. At maximum height, vf  0, so Free fall is the motion of an object v i2 v i2 xM    3 xE when gravity is the only significant force on it. Air significantly affects the  2agrav,M 2 (1/3)agrav,E  paper but not the book. The maximum height would be three times higher on Mars. 47. Final Velocity Your sister drops your b. How would its flight time compare? house keys down to you from the second floor window, as shown in Figure 25. What 1 is the velocity of the keys when you catch x 2 at 2 them? 2 xM 2(3 x E ) tM    3tE Let upward be the positive direction. agrav,M (1/3)agrav,E vf2  vi2  2ax where a  g Flight time is three times longer on Mars.  v i  2g  x 2 vf 50. Velocity and Acceleration Suppose you  (0.0 m/s)  (2)(9.8 m/s )(4.3) 2 2 throw a ball straight up into the air.  9.2 m/s downward Describe the changes in the velocity of the ball. Describe the changes in the 48. Free-Fall Ride Suppose a free-fall ride at acceleration of the ball. an amusement park starts at rest and is in Velocity decreases at a constant rate free fall. What is the velocity of the ride as the ball travels upward. At the after 2.3 s? How far do people on the ride ball’s highest point, velocity is zero. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. fall during the 2.3-s time period? As the ball begins to drop, the velocity Let upward be the positive direction. begins to increase in the negative vf  vi  at direction until it reaches the height  (0 m/s) 9.8 m/s2)(2.3 s) from which it was initially released. At  23 m/s that point, the ball has the same  23 m/s downward speed it had upon release. The 1 acceleration is constant throughout the ball’s flight. xf  xi  vit  2 at2 where xi  0 m and vi  0 m/s, so 51. Critical Thinking A ball thrown vertically 1 1 upward continues upward until it reaches a xf  2 at  2 (9.8 m/s2)(2.3 s)2 2 certain position and then falls downward.  26 m The ball’s velocity is instantaneously zero The people fall 26 m during the 2.3-s at that highest point. Is the ball accelerating time period. at that point? Devise an experiment to prove or disprove your answer. Sample answer: The ball is accelerating; its velocity is changing. Take a multiflash photo to measure its position. From photos, calculate the ball’s velocity. Solutions Manual 10 Inspire Physics  Module 3 continued Go Further Data Analysis Lab How do free fall motion on Earth and Jupiter compare? Suppose a ball could be thrown vertically upward with the same initial velocity on Earth and on Jupiter. An object on the planet Jupiter has about three times the free-fall acceleration as on Earth. Neglect the effects of atmospheric resistance and assume gravity is the only force on the ball. Analyze and Interpret Data 1. Claim How would the maximum height reached by the ball on Jupiter compare to the maximum height reached on Earth? The maximum height reached on Jupiter is 1/3 the height on Earth. 2. Evidence and Reasoning Use sketches and mathematics to explain your approach to this problem and your reasoning to justify your claim. Let J = Jupiter, E = Earth, and agrav = gravitational acceleration. At maximum height, vf = 0, so Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3. How would your claim change if the initial velocity of the ball on Jupiter were three times greater? Explain your reasoning. With vf = 0, the value xf is directly proportional to the square of initial velocity, vi. That is, xf = vf2/(2agrav) (3vi)2/(2agrav). On Earth, an initial velocity three times greater results in a ball rising nine times higher. On Jupiter, however, the height of nine times higher would be reduced to only three times higher because of xf’s inverse relationship to agrav that is three times greater. Inspire Physics 11 Solutions Manual