Solution. ANSWERS - AP Physics Multiple Choice Practice Kinematics. Answer

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1 NSWRS - P Physics Multiple hoice Practice Kinematics Solution nswer 1. Total istance = 60 miles, total time = 1.5 hours; average spee = total istance/total time 2. rea boune by the curve is the isplacement y inspection of particle the positive area between 0 an 1s will be countere by an equal negative area between 1 an 2s. 3. onstant non-zero acceleration woul be a straight line with a non-zero slope 4. rea boune by the curve is the isplacement y inspection of particle the positive area between 0 an 1s will be countere by an equal negative area between 1 an 2s. 5. rea boune by the curve is the isplacement y inspection the negative area between 0 an 1s will be countere by an equal negative area sometime between 1 an 2s. 6. etween 0 an 1 s; 1 = vt; from 1 to 11 secons; 2 = v 0 t + ½ at 2 ; = Since the slope is positive an constant, so is the velocity, therefore the acceleration must be zero 8. For a horizontal projectile, the initial spee oes not affect the time in the air. Use v 0y = 0 with 10 m = ½ gt 2 9. The time in the air for a horizontal projectile is epenent on the height an inepenent of the initial spee. Since the time in the air is the same at spee v an at spee 2v, the istance ( = vt) will be twice as much at a spee of 2v 10. verage velocity = total isplacement/total time; magnitue of total isplacement = 500 m (3-4-5 triangle) an total time = 150 secons 11. The acceleration is constant an negative which means the slope of the velocity time graph must have a constant negative slope. (Only one choice has the correct acceleration anyway) 12. From rest, h = ½ gt t the top of its path, the vertical component of the velocity is zero, which makes the spee at the top a minimum. With symmetry, the projectile has the same spee when at the same height, whether moving up or own. 14. g points own in projectile motion. lways. 15. For a horizontal projectile; h = ½ gt 2 (initial vertical component of velocity is zero) 16. t every point of a projectiles free-fall, the acceleration is the acceleration ue to gravity 17. verage spee = total istance/total time = (8 m 2 m)/(1 secon) 18. For a horizontal projectile; h = ½ gt 2 (initial vertical component of velocity is zero) 19. The area uner the curve is the isplacement. There is more area uner the curve for ar X. 20. rea uner the curve is the isplacement. ar Y is moving faster as they reach the same point. 21. Uniformly accelerate means the spee-time graph shoul be a stright line with non-zero slope. The corresponing istance-time graph shoul have an increasing slope (curve upwar) 22. From the equation = ½ at 2, isplacement is proportional to time square. Traveling from rest for twice the time gives 4 times the isplacement (or 4 m). Since the object alreay travelle 1 m in the first secon, uring the time interval from 1 s to 2 s the object travelle the remaining 3 m 23. To travel straight across the river, the upstream component of the boat s velocity must cancel the current. Since the spee of the current is the same as the spee of the boat, the boat must hea irectly upstream to cancel the current, which leaves no component across the river

2 24. v iy = 200 m/s sin 30º = 100 m/s. t maximum height v y = 0. Use v y 2 = v iy 2 + 2gh 25. cceleration is proportional to v. v = v 2 v 1 = v 2 + ( v 1 ) 26. From a height of 45 m (= ½ gt 2 ) it takes 3 secons to strike the groun. In that time, the ball thrown travele 30 m. v = /t m/s 2 can be thought of as a change in spee of 9.8 m/s per secon. 28. v i = 0 m/s; v f = 30 m/s; t = 6 s; vi vf v 2 t 29. velocity of package relative to observer on groun v pg = v 1 = velocity of package relative to pilot v pp = v 2 = velocity of pilot relative to groun v po = Putting these together into a right triangle yiels v 2 pg + v = v While the object momentarily stops at its peak, it never stops accelerating ownwar. 31. Maximum height of a projectile is foun from v y = 0 at max height an v y 2 = v iy 2 + 2gh an gives h max = v iy 2 /2g = (v i sin ) 2 /2g. Fire straight up, = 90º an we have v i = 2 gh Plugging this initial velocity into the equation for a 45º angle (sin 45º = h new = ( 2gh 1 ) 2 /2g = h/2 2 1 ) gives 32. g points own in projectile motion. lways. 33. horizontal velocity v x remains the same thorughout the flight. g remains the same as well. 34. velocity-time graph represents the slope of the isplacement-time graph. nalyzing the v-t graph shows an increasing slope, then a constant slope, then a ecreasing slope (to zero) 35. For a roppe object: = ½ gt For a horizontal projectile, the initial spee oes not affect the time in the air. Use v 0y = 0 with 10 m = ½ gt 2 to get t = 2 For the horzontal part of the motion; v = /t 37. velocity-time graph represents the slope of the isplacement-time graph. nalyzing the v-t graph shows a constant slope, then a ecreasing slope to zero, becoming negative an increasing, then a constant slope. Note this is an analysis of the values of v, not the slope of the graph itself 38. y process of elimination ( an are unrealistic; is wrong, air resistance shoul ecrease the acceleration; is irrelevant) 39. The 45 angle gives the maximum horizontal travel to the original elevation, but the smaller angle causes the projectile to have a greater horizontal component of velocity, so given the aitional time of travel allows such a trajectory to avance a greater horizontal istance. In other wors given enough time the smaller angle of launch gives a parabola which will eventual cross the parabola of the 45 launch. 40. The area uner the curve of an acceleration-time graph is the change in spee. 41. In the 4 secons to reach the groun, the flare travelle 70 m/s 4 s = 280 m horizontally. 2

3 42. In the 4 secons to reach the groun, the flare travelle 70 m/s 4 s = 280 m horizontally. The plane travelle = v i t + ½ at 2 = (70 m/s)(4 s) + (0.5)(0.75 m/s 2 )(4 s) 2 = 280 m + 6 m, or 6 m ahea of the flare. 43. Positive velocity = positive slope. Negative acceleration = ecreasing slope (or ownwar curvature 44. The slope of the line represents her velocity. eginning positive an constant, going to zero, then positive an larger than the initial, then negative while the line returns to the time axis 45. roppe from a height of 9.8 m, the life preserver takes (9.8 m = ½ gt 2 ); t = 1.4 secons to reach the water. In that 1.4 secons the swimmer covere (6.0 m 2.0 m) = 4.0 m meaning the water spee is (4.0 m)/(1.4 s) 46. The ball takes time T/2 to reach height H. Using v y = 0 at maximum height an vi vf v gives the initial spee as 4H/T. In aition from the top H = ½ g(t/2) 2 2 = t gt 2 /8. Plugging in a time T/4 gives = (4H/T)(T/4) + ½ ( g)(t/4) 2 = H ¼ (gt 2 /8) = ¾ H 47. While the object momentarily stops at its peak, it never stops accelerating ownwar. Without air resistance, symmetry ictates time up = time own. With air resistance consiere, the ball will have a larger average velocity on the way up an a lower average velocity on the way own since it will lan with a smaller spee than it was thrown, meaning the ball takes longer to fall. 48. Total istance =. Time for first ¾ is t 1 = (¾)/v = 3/4v. Time for secon part is t 2 = (¼ )/(½ v) = 2/4v. Total time is then t 1 + t 2 = 5/4v. verage spee = /(5/4v) 49. Positive acceleration is an increasing slope (incluing negative slope increasing towar zero) or upwar curvature 50. With air resistance, the acceleration (the slope of the curve) will ecrease towar zero as the ball reache terminal velocity. Note: without air resistance, choice () woul be correct 51. Since for the first 4 secons, the car is accelerating positively the entire time, the car will be moving fastest just beofre slowing own after t = 4 secons. 52. The area uner the curve represents the change in velocity. The car begins from rest with an increasing positive velocity, after 4 secons the car begins to slow an the area uner the curve from 4 to 8 secons couters the increase in velocity form 0 to 4 secons, bringing the car to rest. However, the car never change irection an was moving away from its original starting position the entire time. 53. The ball will lan with a spee given by the equaion v 2 = v i 2 + 2gH or v = with ¾ the spee gives a new height of v f = 0 = ( ¾ 2 gh ) 2 + 2( g)h new 54. The velocity-time graph shoul represent the slope of the position-time graph an the acceleration-time graph shoul represent the slope of the velocity-time graph 2 gh. Rebouning 55. It s a surprising result, but while both the horizontal an verticla components change at a given height with varying launch angle, the spee (v x 2 + v y 2 ) 1/2 will be inepenent of α (try it!) 56. v f 2 = v i 2 + 2a = (+7 m/s)(2 s) + ½ ( 10 m/s 2 )(2 s) 2 ; 1 = ( 7 m/s)(2 s) + ½ ( 10 m/s 2 )(2 s) 2

4 58. Range of a projectile R = (v i 2 sin 2 )/g an maximum range occurs at = 45º, which gives v i = Rg. Maximum height of a projectile is foun from v y = 0 at max height an v y 2 = v iy 2 + 2gh an gives h max = v iy 2 /2g = (v i sin ) 2 /2g. Maximum range occurs at 45º, which gives h = (Rg)(sin 45) 2 /2g 59. v f 2 = v i 2 + 2a 60. The iagonal of a face of the cube is 3 2m. The iagonal across the cube itself is the hypoteneuse of this face iagonal an a cube ege: ( 3 2) Instantaneous velocity is the slope of the line at that point 62. isplacement is the area uner the curve. Maximum isplacement is just before the car turns aroun at 2.5 secons. 63. Range of a projectile R = (v i 2 sin 2 )/g an maximum range occurs at = 45º, which gives v i = Rg. Using = 30º gives R new = R sin 60º 64. (avance question!) The time for one bounce is foun from v = v + ( g)t which gives t = 2v/g. We are summing the time for all bounces, while the velocity (an hence the time) converge in a geometric series with the ratio v n+1 /v n = r <1 to 1 1 r 65. The acceleration is the slope of the curve at 90 secons. 66. From the equation = ½ at 2, isplacement is proportional to time square. Traveling from rest for twice the time gives 4 times the isplacement (or 4 h). Since the object alreay travelle h in the first secon, uring the time interval from 1 s to 2 s the object travelle the remaining 3h 67. = v i t + ½ gt The relative spee between the coyote an the prairie og is 14.5 m/s. To cover the 45 m istance between them will take t = /v = (45 m)/(14.5 m/s) 69. For the first part of the trip (the thrust): 1 = v i t + ½ at 2 = 0 m + ½ (50 m/s 2 )(2 s) 2 = 100 m For the secon part, we first fin the velocity after the thrust v = at = 100 m/s an at the maximum height v f = 0, so to fin 2 we use v f 2 = v i 2 + 2a 2 which gives 2 = 510 m 70. Total isplacement west = 1100 m; total isplacement south = 400 m. Use the Pythagorean theorem. 71. For a horizontal projectile (v iy = 0 m/s) to fall 1 m takes (using 1 m = ½ gt 2 ) 0.45 secons. To travel 30 m in this time requires a spee of /t = (30 m)/(0.45 s) 72. Maximum height of a projectile is foun from v y = 0 m/s at max height an (0 m/s) 2 = v 2 + 2gh an gives h = v 2 /2g The height at which the projectile is moving with half the spee is foun from (½v) 2 = v 2 + 2( g) which gives = 3v 2 /8g = 0.75 h 73. Looking at choices, an eliminates the possibility of choices an (each ball increases its spee by 9.8 m/s each secon, negating those choices anyway). Since ball is moving faster than ball at all times, it will continue to pull away from ball (the relative spee between the balls separates them). 74. Since they all have the same horizontal component of the shell s velocity, the shell that spens the longest time in the air will travel the farthest. That is the shell launche at the largest angle (mass is irrelevant).

5 75. This is merely asking for the horizontal range of a horizontal projectile. The time in the air is 2 h foun from the height using h = ½ gt 2 which gives t =. The range is foun using = vt g 76. Flying into the win the airliners spee relative to the groun is 500 km/h 100 km/h = 400 km/h an a 3000 km trip will take t = /v = 7.5 hours. Flying with the win the airliners spee relative to the groun is 500 km/h km/h = 600 km/h an a 3000 km trip will take t = /v = 5 hours making the total time 12.5 hours. 77. The horizontal component of the velocity is 28.3 m/s cos 60º = m/s. If the ball is in the air for 5 secons the horizontal isplacement is x = v x t 78. since (from rest) = ½ gt 2, istance is proportional to time square. n object falling for twice the time will fall four times the istance. 79. vi vf v 2 t 80. v f = 40 m/s (negative since it is moving own when laning). Use v f = v i + ( g)t 81. For a horizontal projectile (v iy = 0 m/s) to fall 0.05 m takes (using 0.05 m = ½ gt 2 ) 0.1 secons. To travel 20 m in this time requires a spee of /t = (20 m)/(0.1 s) m/s 2 means the spee changes by 9.8 m/s each secon 83. Once release, the package is in free-fall (subject to gravity only) 84. For the first part v = at = 8.0 m/s an = ½ at 2 = 40 m. In the secon part of the trip, the spee remains at 8 m/s, an travels an aitional = vt = 80 m 85. The efinition of termincal velocity is the velocity at which the force of air friction balances the weight of the object an the object no longer accelerates. 86. To reach a spee of 30 m/s when roppe takes (using v = at) about 3 secons. The istance fallen after three secons is foun using = ½ at m/s 2 means the spee changes by 9.8 m/s each secon (in the ownwar irection) 88. Total istance = 800 km. Times are (400 km)/(80 km/h) = 5 hours an (400 km)/(100 km/h) = 4 hours. verage spee = total istance ivie by total time. 89. v = at m/s 2 means the spee changes by 9.8 m/s each secon 91. Velocity is a vector, spee is a scalar 92. hoices,, an all refer to vectors 93. Falling on the Moon is no ifferent conceptually than falling on the arth 94. Since the line is above the t axis for the entire flight, the uck is always moving in the positive (forwar) irection, until it stops at point 95. One coul analyze the graphs base on slope, but more simply, the graph of position versus time shoul represent the actual path followe by the ball as seen on a platform moving past you at constant spee. 96. Other than the falling portions (a = 9.8 m/s 2 ) the ball shoul have a spike in the acceleration when it bounces ue to the rapi change of velocity from ownwar to upwar. 97. The same average spee woul be inicate by the same istance travelle in the time interval

6 98. t t 3, car #1 is ahea of car #2 an at t 4, car #1 is behin car #2. They were in the same position somewhere in between 99. verage spee = (total istance)/(total time). ars #2 an #3 travelle the same istance If you look at the istance covere in each time interval you shoul nitce a pattern: 2 m, 6 m, 10 m, 14 m, 18 m; making the istance in the next secon 22 m Instantaneous spee is the slope of the line at that point non-zero accleeration is inicate by a curve in the line 103. Net isplacement north = 300 miles sin 30º = 150 miles Net isplacement east = (300 miles cos 30º 600 miles) = 340 miles, or 340 miles west miles 340miles ngle north of west is tan ( ) 104. Maximum height of a projectile is foun from v y = 0 m/s at max height an (0 m/s) 2 = v 2 + 2gh an gives h = v 2 /2g. t twice the initial spee, the height will be 4 times as much 105. verage spee = total istance ivie by total time = (0.48 m)/(0.2 s) 106. = ½ at 2 (use any point) 107. v = v i + at 108. cceleration is the slope of the line segment 109. isplacement is the area uner the line 110. v i = 30 m, v f = 0, = 45 m; vi vf v 2 t 111. In a vacuum, there is no air resistance an hence no terminal velocity. It will continue to accelerate projectile launche at a smaller angle oes not go as high an will fall to the groun first v x = v i cos 114. Velocity is the slope of the line Positive acceleration is an upwar curvature 116. verage acceleration = v/ t 117. = ½ at cceleration is the slope of the line segment 119. isplacement is the area between the line an the t-axis. rea is negative when the line is below the t-axis fter two secons, the object woul be above it s original position, still moving upwar, but the acceleration ue to gravity is always pointing own 121. onstant spee is a constant slope on a position-time graph, a horizontal line on a velocity time graph or a zero value on an acceleration-time graph 122. verage spee = total istance ivie by total time = (7 cm)/(1 s) 123. = ½ at 2 (use any point)

7 124. Maximum height of a projectile is foun from v y = 0 m/s at max height an (0 m/s) 2 = v 2 + 2gh an gives h = v 2 /2g. Mass is irrelevant. Largest initial spee = highest Using = ½ at 2 shows the height is proportional to the time square. ½ the maximum height is 1 times the time Stopping istance is foun using v f = 0 = v i 2 + 2a which gives = v i 2 /2a where stopping istance is proportional to initial spee square v f = v i + gt 128. Moving away from the origin will maintain a negative position an velocity. Slowing own inicates the acceleration is opposite in irection to the velocity The arrow travels equal horizontal istances in equal amounts of time. The istance fallen is proportional to time square. The arrow will have fallen a total of 0.8 m in the next 5 m horizontally, or an aitional 0.6 m Tan 53º = h/(8 m) 131. = ½ at Maximum height of a projectile is foun from v y = 0 at max height an v y 2 = v iy 2 + 2gh an gives h max = v iy 2 /2g = (v i sin ) 2 /2g 133. cceleration is the slope of the line 134. Since the first rock is always traveling faster, the relative istance between them is always increasing Stopping istance is foun using v f = 0 = v i 2 + 2a which gives = v i 2 /2a where stopping istance is proportional to initial spee square t an angle of 120º, there is a component of the acceleration perpenicular to the velocity causing the irection to change an a component in the opposite irection of the velocity, causing it to slow own = ½ at The isplacement is irectly to the left. The average velocity is proportional to the isplacement 139. The velocity is initially pointing up, the final velocity points own. The acceleration is in the same irection as v = v f + ( v i ) 140. The car is the greatest istance just before it reverses irection at 5 secons verage spee = (total istance)/(total time), the total istance is the magnitue of the area uner the line (the area below the t-axis is consiere positive) 142. Spee is the slope of the line velocity is pointing tangent to the path, acceleration (gravity) is ownwar verage spee = (total istance)/(total time) 145. To travel 120 m horizontally in 4 s gives v x = 30 m/s. The time to reach maximum height was 2 secons an v y = 0 at the maximum height which gives v iy = 20 m/s. v i = 2 2 vx v iy 146. The relative spee between the two cars is v 1 v 2 = (60 km/h) ( 40 km/h) = 100 km/h. They will meet in t = /v relative = 150 km/100 km/h

8 147. cceleration is inepenent of velocity (you can accelerate in any irection while traveling in a ny irection) /4 =3, now the units: M = 10 6, T = : M/T = 10 6 = micro ( ) 149. cceleration is inepenent of velocity (you can accelerate in any irection while traveling in a ny irection). If the accleration is in the same irection as the velocity, the object is speeing up s the first bales roppe will always be traveling faster than the later bales, their relative velocity will cause their separation to always iincrease Horizontally, the bales will all travel at the spee of the plane, as gravity will not affect their horizontal motion. = vt = (50 m/s)(2 secons apart) 152. Traveling in still water will take a time t = /v = 2/v. Traveling perpenicularly across the stream requires the boat to hea at an angle into the current, causing the relative velocity of the boat to the shore to be less than when in still water an therefore take a longer time. Since this eliminates choice an choices an are ientical, that leaves as the only single option. If you really want proof: To show an take longer, we have the following (let the current be moving with spee w): v w v w 2v 2 2 v w v w v2 w2 w2 v v v traveling ownstream; v rel = v + w an time = traveling upstream; v rel = v w an time = total time = 153. When the first car starts the last lap, it will finish the race in 15 secons from that point. In 15 secons, the secon car will travel (1 km/12 s) 15 s = 1250 m so the first car must be at least 250 m ahea when starting the last lap to win the race.