Física Young & Freedman Solutions Capítulo 2 14ed, Exercícios de Física

Baixe Física Young & Freedman Solutions Capítulo 2 14ed e outras Exercícios em PDF para Física, somente na Docsity! © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2-1 2.1. IDENTIFY: av-xx v tΔ = Δ SET UP: We know the average velocity is 6.25 m/s. EXECUTE: av- 25 0 mxx v tΔ = Δ = . EVALUATE: In round numbers, 6 m/s × 4 s = 24 m ≈ 25 m, so the answer is reasonable. 2.2. IDENTIFY: av-x xv t Δ= Δ SET UP: 613 5 days 1 166 10 s.. = . × At the release point, 65 150 10 m.x = + . × EXECUTE: (a) 6 2 1 av- 6 5 150 10 m 4 42 m/s. 1 166 10 sx x xv t − − . ×= = = − . Δ . × (b) For the round trip, 2 1x x= and 0.xΔ = The average velocity is zero. EVALUATE: The average velocity for the trip from the nest to the release point is positive. 2.3. IDENTIFY: Target variable is the time tΔ it takes to make the trip in heavy traffic. Use Eq. (2.2) that relates the average velocity to the displacement and average time. SET UP: av-x xv t Δ= Δ so av-xx v tΔ = Δ and av-x xt v ΔΔ = . EXECUTE: Use the information given for normal driving conditions to calculate the distance between the two cities, where the time is 1 h and 50 min, which is 110 min: av- (105 km/h)(1 h/60 min)(110 min) 192.5 kmxx v tΔ = Δ = = . Now use av-xv for heavy traffic to calculate ;tΔ xΔ is the same as before: av- 192.5 km 2.75 h 2 h 70 km/hx xt v ΔΔ = = = = and 45 min. The additional time is (2 h and 45 min) – (1 h and 50 min) = (1 h and 105 min) – (1 h and 50 min) = 55 min. EVALUATE: At the normal speed of 105 km/s the trip takes 110 min, but at the reduced speed of 70 km/h it takes 165 min. So decreasing your average speed by about 30% adds 55 min to the time, which is 50% of 110 min. Thus a 30% reduction in speed leads to a 50% increase in travel time. This result (perhaps surprising) occurs because the time interval is inversely proportional to the average speed, not directly proportional to it. 2.4. IDENTIFY: The average velocity is av- .x xv t Δ= Δ Use the average speed for each segment to find the time traveled in that segment. The average speed is the distance traveled divided by the time. SET UP: The post is 80 m west of the pillar. The total distance traveled is 200 m 280 m 480 m.+ = EXECUTE: (a) The eastward run takes time 200 m 40 0 s 5 0 m/s = . . and the westward run takes 280 m 70 0 s. 4 0 m/s = . . The average speed for the entire trip is 480 m 4 4 m/s. 110 0 s = . . (b) av- 80 m 0 73 m/s. 110 0 sx xv t Δ −= = = − . Δ . The average velocity is directed westward. MOTION ALONG A STRAIGHT LINE 2 2-2 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EVALUATE: The displacement is much less than the distance traveled, and the magnitude of the average velocity is much less than the average speed. The average speed for the entire trip has a value that lies between the average speed for the two segments. 2.5. IDENTIFY: Given two displacements, we want the average velocity and the average speed. SET UP: The average velocity is av-x xv t Δ= Δ and the average speed is just the total distance walked divided by the total time to walk this distance. EXECUTE: (a) Let +x be east. 60 0 m 40 0 m 20 0 mxΔ = . − . = . and 28 0 s 36 0 s 64 0 s.tΔ = . + . = . So av- 20 0 m 0 312 m/s. 64 0 sx xv t Δ .= = = . Δ . (b) 60 0 m 40 0 maverage speed 1 56 m/s 64 0 s . + .= = . . EVALUATE: The average speed is much greater than the average velocity because the total distance walked is much greater than the magnitude of the displacement vector. 2.6. IDENTIFY: The average velocity is av- .x xv t Δ= Δ Use ( )x t to find x for each t. SET UP: (0) 0,x = (2 00 s) 5 60 m,x . = . and (4 00 s) 20 8 mx . = . EXECUTE: (a) av- 5 60 m 0 2 80 m/s 2 00 sx v . −= = + . . (b) av- 20 8 m 0 5 20 m/s 4 00 sx v . −= = + . . (c) av- 20 8 m 5 60 m 7 60 m/s 2 00 sx v . − .= = + . . EVALUATE: The average velocity depends on the time interval being considered. 2.7. (a) IDENTIFY: Calculate the average velocity using av- .x xv t Δ= Δ SET UP: av-x xv t Δ= Δ so use ( )x t to find the displacement xΔ for this time interval. EXECUTE: 0 :t = 0x = 10 0 s:t = . 2 2 3 3(2 40 m/s )(10 0 s) (0 120 m/s )(10 0 s) 240 m 120 m 120 mx = . . − . . = − = . Then av- 120 m 12 0 m/s 10 0 sx xv t Δ= = = . . Δ . (b) IDENTIFY: Use x dxv dt = to calculate ( )xv t and evaluate this expression at each specified t. SET UP: 22 3x dxv bt ct dt = = − . EXECUTE: (i) 0 :t = 0xv = (ii) 5 0 s:t = . 2 3 22(2 40 m/s )(5 0 s) 3(0 120 m/s )(5 0 s) 24 0 m/s 9 0 m/s 15 0 m/sxv = . . − . . = . − . = . . (iii) 10 0 s:t = . 2 3 22(2 40 m/s )(10 0 s) 3(0 120 m/s )(10 0 s) 48 0 m/s 36 0 m/s 12 0 m/sxv = . . − . . = . − . = . . (c) IDENTIFY: Find the value of t when ( )xv t from part (b) is zero. SET UP: 22 3xv bt ct= − 0xv = at 0t = . 0xv = next when 22 3 0bt ct− = EXECUTE: 2 3b ct= so 2 3 2 2(2 40 m/s ) 13 3 s 3 3(0 120 m/s ) bt c .= = = . . EVALUATE: ( )xv t for this motion says the car starts from rest, speeds up, and then slows down again. Motion Along a Straight Line 2-5 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Set 60 0 cmx = . and solve for t: 6 20 st = . and 25 8 st = . . At 6 20 s,t = . 1 23 cm/sxv = + . . At 25 8 s,t = . 1 23 cm/sxv = − . . Set 40 0 cmx = . and solve for t: 36 4 st = . (other root to the quadratic equation is negative and hence nonphysical). At 36 4 s,t = . 2 55 cm/sxv = − . . (e) The graphs are sketched in Figure 2.15. Figure 2.15 EVALUATE: The acceleration is constant and negative. xv is linear in time. It is initially positive, decreases to zero, and then becomes negative with increasing magnitude. The turtle initially moves farther away from the origin but then stops and moves in the -directionx− . 2.16. IDENTIFY: Use av- xx va t Δ= Δ , with 10 stΔ = in all cases. SET UP: xv is negative if the motion is to the left. EXECUTE: (a) 2[(5 0 m/s) (15 0 m/s)]/(10 s) 1 0 m/s. − . = − . (b) 2[( 15 0 m/s) ( 5 0 m/s)]/(10 s) 1 0 m/s− . − − . = − . (c) 2[( 15 0 m/s) ( 15 0 m/s)]/(10 s) 3 0 m/s− . − + . = − . EVALUATE: In all cases, the negative acceleration indicates an acceleration to the left. 2.17. IDENTIFY: The average acceleration is av- .xx va t Δ= Δ Use ( )xv t to find xv at each t. The instantaneous acceleration is .xx dva dt = SET UP: (0) 3 00 m/sxv = . and (5 00 s) 5 50 m/s.xv . = . EXECUTE: (a) 2av- 5 50 m/s 3 00 m/s 0 500 m/s 5 00 s x x va t Δ . − .= = = . Δ . (b) 3 3(0 100 m/s )(2 ) (0 200 m/s ) .xx dva t t dt = = . = . At 0,t = 0.xa = At 5 00 s,t = . 21 00 m/s .xa = . (c) Graphs of ( )xv t and ( )xa t are given in Figure 2.17 (next page). EVALUATE: ( )xa t is the slope of ( )xv t and increases as t increases. The average acceleration for 0t = to 5 00 st = . equals the instantaneous acceleration at the midpoint of the time interval, 2 50 s,t = . since ( )xa t is a linear function of t. 2-6 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Figure 2.17 2.18. IDENTIFY: ( )x dxv t dt = and ( ) xx dva t dt = SET UP: 1( )n nd t nt dt −= for 1.n ≥ EXECUTE: (a) 2 6 5( ) (9 60 m/s ) (0 600 m/s )xv t t t= . − . and 2 6 4( ) 9 60 m/s (3 00 m/s ) .xa t t= . − . Setting 0xv = gives 0t = and 2 00 s.t = . At 0,t = 2 17 mx = . and 29 60 m/s .xa = . At 2 00 s,t = . 15 0 mx = . and 238 4 m/s .xa = − . (b) The graphs are given in Figure 2.18. EVALUATE: For the entire time interval from 0t = to 2 00 s,t = . the velocity xv is positive and x increases. While xa is also positive the speed increases and while xa is negative the speed decreases. Figure 2.18 2.19. IDENTIFY: Use the constant acceleration equations to find 0xv and xa . (a) SET UP: The situation is sketched in Figure 2.19. 0 70 0 mx x− = . 6 00 st = . 15.0 m/sxv = 0 ?xv = Figure 2.19 Motion Along a Straight Line 2-7 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EXECUTE: Use 00 ,2 x xv vx x t+⎛ ⎞− = ⎜ ⎟ ⎝ ⎠ so 00 2( ) 2(70 0 m) 15 0 m/s 8.33 m/s 6 00 sx x x xv v t − .= − = − . = . . (b) Use 0 ,x x xv v a t= + so 20 15 0 m/s 5 0 m/s 1 11 m/s 6 00 s x x x v va t − . − .= = = . . . EVALUATE: The average velocity is (70 0 m)/(6 00 s) 11.7 m/s. . = . The final velocity is larger than this, so the antelope must be speeding up during the time interval; 0x xv v< and 0xa > . 2.20. IDENTIFY: In (a) find the time to reach the speed of sound with an acceleration of 5g, and in (b) find his speed at the end of 5.0 s if he has an acceleration of 5g. SET UP: Let x+ be in his direction of motion and assume constant acceleration of 5g so the standard kinematics equations apply so 0 .x x xv v a t= + (a) 3(331 m/s) 993 m/s,xv = = 0 0,xv = and 25 49 0 m/sxa g= = . . (b) 5 0 st = . EXECUTE: (a) 0x x xv v a t= + and 0 2 993 m/s 0 20 3 s. 49 0 m/s x x x v vt a − − = = = . . Yes, the time required is larger than 5.0 s. (b) 20 0 (49 0 m/s )(5 0 s) 245 m/s.x x xv v a t= + = + . . = EVALUATE: In 5.0 s he can only reach about 2/3 the speed of sound without blacking out. 2.21. IDENTIFY: For constant acceleration, the standard kinematics equations apply. SET UP: Assume the ball starts from rest and moves in the -directionx+ . EXECUTE: (a) 0 1 50 m,x x− = . 45 0 m/sxv = . and 0 0.xv = 2 2 0 02 ( )x x xv v a x x= + − gives 2 2 2 20 0 (45 0 m/s) 675 m/s . 2( ) 2(1 50 m) x x x v va x x − .= = = − . (b) 00 2 x xv vx x t+⎛ ⎞− = ⎜ ⎟ ⎝ ⎠ gives 0 0 2( ) 2(1 50 m) 0 0667 s 45 0 m/sx x x xt v v − .= = = . + . EVALUATE: We could also use 0x x xv v a t= + to find 2 45 0 m/s 0 0667 s 675 m/s x x vt a .= = = . which agrees with our previous result. The acceleration of the ball is very large. 2.22. IDENTIFY: For constant acceleration, the standard kinematics equations apply. SET UP: Assume the ball moves in the directionx+ . EXECUTE: (a) 73 14 m/s,xv = . 0 0xv = and 30 0 ms.t = . 0x x xv v a t= + gives 20 3 73 14 m/s 0 2440 m/s . 30 0 10 s x x x v va t − − . −= = = . × (b) 300 0 73 14 m/s (30 0 10 s) 1 10 m 2 2 x xv vx x t −+ + .⎛ ⎞ ⎛ ⎞− = = . × = .⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . EVALUATE: We could also use 210 0 2x xx x v t a t− = + to calculate 0:x x− 2 3 21 0 2 (2440 m/s )(30 0 10 s) 1 10 m,x x −− = . × = . which agrees with our previous result. The acceleration of the ball is very large. 2.23. IDENTIFY: Assume that the acceleration is constant and apply the constant acceleration kinematic equations. Set | |xa equal to its maximum allowed value. SET UP: Let x+ be the direction of the initial velocity of the car. 2250 m/s .xa = − 105 km/h 29 17 m/s.= . EXECUTE: 0 29 17 m/s.xv = . 0.xv = 2 2 0 02 ( )x x xv v a x x= + − gives 2 2 2 0 0 2 0 (29 17 m/s) 1 70 m. 2 2( 250 m/s ) x x x v vx x a − − .− = = = . − EVALUATE: The car frame stops over a shorter distance and has a larger magnitude of acceleration. Part of your 1.70 m stopping distance is the stopping distance of the car and part is how far you move relative to the car while stopping. 2-10 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.31. (a) IDENTIFY and SET UP: The acceleration xa at time t is the slope of the tangent to the xv versus t curve at time t. EXECUTE: At 3 s,t = the xv versus t curve is a horizontal straight line, with zero slope. Thus 0xa = . At 7 s,t = the xv versus t curve is a straight-line segment with slope 245 m/s 20 m/s 6 3 m/s 9 s 5 s − = . . − Thus 26 3 m/sxa = . . At 11 st = the curve is again a straight-line segment, now with slope 20 45 m/s 11 2 m/s 13 s 9 s − − = − . . − Thus 211 2 m/sxa = − . . EVALUATE: 0xa = when xv is constant, 0xa > when xv is positive and the speed is increasing, and 0xa < when xv is positive and the speed is decreasing. (b) IDENTIFY: Calculate the displacement during the specified time interval. SET UP: We can use the constant acceleration equations only for time intervals during which the acceleration is constant. If necessary, break the motion up into constant acceleration segments and apply the constant acceleration equations for each segment. For the time interval 0t = to 5 st = the acceleration is constant and equal to zero. For the time interval 5 st = to 9 st = the acceleration is constant and equal to 26 25 m/s. . For the interval 9 st = to 13 st = the acceleration is constant and equal to 211 2 m/s− . . EXECUTE: During the first 5 seconds the acceleration is constant, so the constant acceleration kinematic formulas can be used. 0 20 m/sxv = 0xa = 5 st = 0 ?x x− = 0 0xx x v t− = ( 0xa = so no 21 2 xa t term) 0 (20 m/s)(5 s) 100 m;x x− = = this is the distance the officer travels in the first 5 seconds. During the interval 5 st = to 9 s the acceleration is again constant. The constant acceleration formulas can be applied to this 4-second interval. It is convenient to restart our clock so the interval starts at time 0t = and ends at time 4 st = . (Note that the acceleration is not constant over the entire 0t = to 9 st = interval.) 0 20 m/sxv = 26 25 m/sxa = . 4 st = 0 100 mx = 0 ?x x− = 21 0 0 2x xx x v t a t− = + 2 21 0 2(20 m/s)(4 s) (6 25 m/s )(4 s) 80 m 50 m 130 m.x x− = + . = + = Thus 0 130 m 100 m 130 m 230 mx x− + = + = . At 9 st = the officer is at 230 m,x = so she has traveled 230 m in the first 9 seconds. During the interval 9 st = to 13 st = the acceleration is again constant. The constant acceleration formulas can be applied for this 4-second interval but not for the whole 0t = to 13 st = interval. To use the equations restart our clock so this interval begins at time 0t = and ends at time 4 st = . 0 45 m/sxv = (at the start of this time interval) 211 2 m/sxa = − . 4 st = 0 230 mx = 0 ?x x− = 21 0 0 2x xx x v t a t− = + 2 21 0 2(45 m/s)(4 s) ( 11 2 m/s )(4 s) 180 m 89 6 m 90 4 m.x x− = + − . = − . = . Thus 0 90 4 m 230 m 90 4 m 320 mx x= + . = + . = . At 13 st = the officer is at 320 m,x = so she has traveled 320 m in the first 13 seconds. EVALUATE: The velocity xv is always positive so the displacement is always positive and displacement and distance traveled are the same. The average velocity for time interval tΔ is av- /xv x t= Δ Δ . For 0t = to 5 s, av- 20 m/sxv = . For 0t = to 9 s, av- 26 m/sxv = . For 0t = to 13 s, av- 25 m/sxv = . These results are consistent with the figure in the textbook. 2.32. IDENTIFY: ( )xv t is the slope of the x versus t graph. Car B moves with constant speed and zero acceleration. Car A moves with positive acceleration; assume the acceleration is constant. Motion Along a Straight Line 2-11 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. SET UP: For car B, xv is positive and 0.xa = For car A, xa is positive and xv increases with t. EXECUTE: (a) The motion diagrams for the cars are given in Figure 2.32a. (b) The two cars have the same position at times when their x-t graphs cross. The figure in the problem shows this occurs at approximately 1 st = and 3 s.t = (c) The graphs of xv versus t for each car are sketched in Figure 2.32b. (d) The cars have the same velocity when their x-t graphs have the same slope. This occurs at approximately 2 s.t = (e) Car A passes car B when Ax moves above Bx in the x-t graph. This happens at 3 s.t = (f) Car B passes car A when Bx moves above Ax in the x-t graph. This happens at 1 s.t = EVALUATE: When 0,xa = the graph of xv versus t is a horizontal line. When xa is positive, the graph of xv versus t is a straight line with positive slope. Figure 2.32 2.33. IDENTIFY: For constant acceleration, the kinematics formulas apply. We can use the total displacement ande final velocity to calculate the acceleration and then use the acceleration and shorter distance to find the speed. SET UP: Take +x to be down the incline, so the motion is in the +x direction. The formula 2 2 0 02 ( )x xv v a x x= + − applies. EXECUTE: First look at the motion over 6.80 m. We use the following numbers: v0x = 0, x – x0 = 6.80 m, and vx = 3.80 /s. Solving the above equation for ax gives ax = 1.062 m/s2. Now look at the motion over the 3.40 m using v0x = 0, ax = 1.062 m/s2 and x – x0 = 3.40 m. Solving the same equation, but this time for vx, gives vx = 2.69 m/s. EVALUATE: Even though the block has traveled half way down the incline, its speed is not half of its speed at the bottom. 2.34. IDENTIFY: Apply the constant acceleration equations to the motion of each vehicle. The truck passes the car when they are at the same x at the same 0.t > SET UP: The truck has 0.xa = The car has 0 0.xv = Let x+ be in the direction of motion of the vehicles. Both vehicles start at 0 0.x = The car has 2 C 2.80 m/s .a = The truck has 20 0 m/s.xv = . EXECUTE: (a) 210 0 2x xx x v t a t− = + gives T 0Tx v t= and 21 C C2 .x a t= Setting T Cx x= gives 0t = and 1 0T C2 ,v a t= so 0T 2 C 2 2(20 0 m/s) 14.29 s. 2.80 m/s vt a .= = = At this t, T (20 0 m/s)(14.29 s) 286 mx = . = and 2 21 2 (3 20 m/s )(14.29 s) 286 m.x = . = The car and truck have each traveled 286 m. (b) At 14.29 s,t = the car has 20 (2.80 m/s )(14.29 s) 40 m/s.x x xv v a t= + = = (c) T 0Tx v t= and 21 C C2 .x a t= The x-t graph of the motion for each vehicle is sketched in Figure 2.34a. (d) T 0T.v v= C C .v a t= The -xv t graph for each vehicle is sketched in Figure 2.34b (next page). EVALUATE: When the car overtakes the truck its speed is twice that of the truck. 2-12 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Figure 2.34 2.35. IDENTIFY: Apply the constant acceleration equations to the motion of the flea. After the flea leaves the ground, ,ya g= downward. Take the origin at the ground and the positive direction to be upward. (a) SET UP: At the maximum height 0yv = . 0yv = 0 0 440 my y− = . 29 80 m/sya = − . 0 ?yv = 2 2 0 02 ( )y y yv v a y y= + − EXECUTE: 20 02 ( ) 2( 9 80 m/s )(0 440 m) 2 94 m/sy yv a y y= − − = − − . . = . (b) SET UP: When the flea has returned to the ground 0 0y y− = . 0 0y y− = 0 2 94 m/syv = + . 29 80 m/sya = − . ?t = 21 0 0 2y yy y v t a t− = + EXECUTE: With 0 0y y− = this gives 0 2 2 2(2 94 m/s) 0 600 s 9 80 m/s y y v t a .= − = − = . . − . EVALUATE: We can use 0y y yv v a t= + to show that with 0 2 94 m/s,yv = . 0yv = after 0.300 s. 2.36. IDENTIFY: The rock has a constant downward acceleration of 9.80 m/s2. We know its initial velocity and position and its final position. SET UP: We can use the kinematics formulas for constant acceleration. EXECUTE: (a) 0 30 m,y y− = − 0 22 0 m/s,yv = . 29 80 m/s .ya = − . The kinematics formulas give 2 2 2 0 02 ( ) (22 0 m/s) 2( 9 80 m/s )( 30 m) 32.74 m/s,y y yv v a y y= − + − = − . + − . − = − so the speed is 32.7 m/s. (b) 0y y yv v a t= + and 0 2 32.74 m/s 22 0 m/s 5.59 s. 9 80 m/s y y y v v t a − − − .= = = − . EVALUATE: The vertical velocity in part (a) is negative because the rock is moving downward, but the speed is always positive. The 5.59 s is the total time in the air. 2.37. IDENTIFY: The pin has a constant downward acceleration of 9.80 m/s2 and returns to its initial position. SET UP: We can use the kinematics formulas for constant acceleration. EXECUTE: The kinematics formulas give 20 0 1 . 2y y y y v t a t− = + We know that 0 0,y y− = so 0 2 2 2(8 20 m/s) 1 67 s. 9 80 m/s y y v t a .= − = − = + . − . EVALUATE: It takes the pin half this time to reach its highest point and the remainder of the time to return. 2.38. IDENTIFY: The putty has a constant downward acceleration of 9.80 m/s2. We know the initial velocity of the putty and the distance it travels. Motion Along a Straight Line 2-15 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EVALUATE: In part (b) we could have found the time from 210 0 2 ,y yy y v t a t− = + finding yv first allows us to avoid solving for t from a quadratic equation. Figure 2.43 2.44. IDENTIFY: Apply constant acceleration equations to the vertical motion of the sandbag. SET UP: Take y+ upward. 29 80 m/s .ya = − . The initial velocity of the sandbag equals the velocity of the balloon, so 0 5 00 m/s.yv = + . When the balloon reaches the ground, 0 40 0 m.y y− = − . At its maximum height the sandbag has 0.yv = EXECUTE: (a) 0 250 s:t = . 2 2 21 10 0 2 2(5 00 m/s)(0 250 s) ( 9 80 m/s )(0 250 s) 0 94 m.y yy y v t a t− = + = . . + − . . = . The sandbag is 40.9 m above the ground. 20 5 00 m/s ( 9 80 m/s )(0 250 s) 2 55 m/s.y y yv v a t= + = + . + − . . = . 1 00 s:t = . 2 210 2(5 00 m/s)(1 00 s) ( 9 80 m/s )(1 00 s) 0 10 m.y y− = . . + − . . = . The sandbag is 40.1 m above the ground. 20 5 00 m/s ( 9 80 m/s )(1 00 s) 4 80 m/s.y y yv v a t= + = + . + − . . = − . (b) 0 40 0 m,y y− = − . 0 5 00 m/s,yv = . 29 80 m/s .ya = − . 21 0 0 2y yy y v t a t− = + gives 2 240 0 m (5 00 m/s) (4 90 m/s ) .t t− . = . − . 2 2(4 90 m/s ) (5 00 m/s) 40 0 m 0t t. − . − . = and ( )21 5 00 ( 5 00) 4(4 90)( 40 0) s (0 51 2 90) s.9 80t = . ± − . − . − . = . ± .. t must be positive, so 3 41 s.t = . (c) 20 5 00 m/s ( 9 80 m/s )(3 41 s) 28 4 m/sy y yv v a t= + = + . + − . . = − . (d) 0 5 00 m/s,yv = . 29 80 m/s ,ya = − . 0.yv = 2 2 0 02 ( )y y yv v a y y= + − gives 2 2 2 0 0 2 0 (5 00 m/s) 1 28 m. 2 2( 9 80 m/s ) y y y v v y y a − − .− = = = . − . The maximum height is 41.3 m above the ground. (e) The graphs of ,ya ,yv and y versus t are given in Figure 2.44. Take 0y = at the ground. EVALUATE: The sandbag initially travels upward with decreasing velocity and then moves downward with increasing speed. Figure 2.44 2-16 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.45. IDENTIFY: Use the constant acceleration equations to calculate xa and 0x x− . (a) SET UP: 224 m/s,xv = 0 0,xv = 0 900 s,t = . ?xa = 0x x xv v a t= + EXECUTE: 20 224 m/s 0 249 m/s 0 900 s x x x v va t − −= = = . (b) 2 2/ (249 m/s )/(9 80 m/s ) 25 4xa g = . = . (c) 2 2 21 10 0 2 20 (249 m/s )(0 900 s) 101 mx xx x v t a t− = + = + . = (d) SET UP: Calculate the acceleration, assuming it is constant: 1 40 s,t = . 0 283 m/s,xv = 0xv = (stops), ?xa = 0x x xv v a t= + EXECUTE: 20 0 283 m/s 202 m/s 1 40 s x x x v va t − −= = = − . 2 2/ ( 202 m/s )/(9 80 m/s ) 20 6;xa g = − . = − . 20 6xa g= − . If the acceleration while the sled is stopping is constant then the magnitude of the acceleration is only 20.6g. But if the acceleration is not constant it is certainly possible that at some point the instantaneous acceleration could be as large as 40g. EVALUATE: It is reasonable that for this motion the acceleration is much larger than g. 2.46. IDENTIFY: Since air resistance is ignored, the egg is in free-fall and has a constant downward acceleration of magnitude 29 80 m/s .. Apply the constant acceleration equations to the motion of the egg. SET UP: Take y+ to be upward. At the maximum height, 0.yv = EXECUTE: (a) 0 30 0 m,y y− = − . 5 00 s,t = . 29 80 m/s .ya = − . 21 0 0 2y yy y v t a t− = + gives 20 1 1 0 2 2 30 0 m ( 9 80 m/s )(5 00 s) 18 5 m/s. 5 00 sy y y yv a t t − − .= − = − − . . = + . . (b) 0 18 5 m/s,yv = + . 0yv = (at the maximum height), 29 80 m/s .ya = − . 2 2 0 02 ( )y y yv v a y y= + − gives 2 2 2 0 0 2 0 (18 5 m/s) 17 5 m. 2 2( 9 80 m/s ) y y y v v y y a − − .− = = = . − . (c) At the maximum height 0.yv = (d) The acceleration is constant and equal to 29 80 m/s ,. downward, at all points in the motion, including at the maximum height. (e) The graphs are sketched in Figure 2.46. EVALUATE: The time for the egg to reach its maximum height is 0 2 18 5 m/s 1 89 s. 9 8 m/s y y y v v t a − − .= = = . − . The egg has returned to the level of the cornice after 3.78 s and after 5.00 s it has traveled downward from the cornice for 1.22 s. Figure 2.46 Motion Along a Straight Line 2-17 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 2.47. IDENTIFY: We can avoid solving for the common height by considering the relation between height, time of fall, and acceleration due to gravity, and setting up a ratio involving time of fall and acceleration due to gravity. SET UP: Let Eng be the acceleration due to gravity on Enceladus and let g be this quantity on earth. Let h be the common height from which the object is dropped. Let y+ be downward, so 0 .y y h− = 0 0yv = EXECUTE: 210 0 2y yy y v t a t− = + gives 21 E2h gt= and 21 En En2 .h g t= Combining these two equations gives 2 2 E En Engt g t= and 2 2 2 2E En En 1 75 s(9 80 m/s ) 0 0868 m/s . 18 6 s tg g t ⎛ ⎞ .⎛ ⎞= = . = .⎜ ⎟ ⎜ ⎟.⎝ ⎠⎝ ⎠ EVALUATE: The acceleration due to gravity is inversely proportional to the square of the time of fall. 2.48. IDENTIFY: Since air resistance is ignored, the boulder is in free-fall and has a constant downward acceleration of magnitude 29 80 m/s .. Apply the constant acceleration equations to the motion of the boulder. SET UP: Take y+ to be upward. EXECUTE: (a) 0 40 0 m/s,yv = + . 20 0 m/s,yv = + . 29 80 m/s .ya = − . 0y y yv v a t= + gives 0 2 20 0 m/s 40 0 m/s 2 04 s. 9 80 m/s y y y v v t a − . − .= = = + . − . (b) 20 0 m/s.yv = − . 0 2 20 0 m/s 40 0 m/s 6 12 s. 9 80 m/s y y y v v t a − − . − .= = = + . − . (c) 0 0,y y− = 0 40 0 m/s,yv = + . 29 80 m/s .ya = − . 21 0 0 2y yy y v t a t− = + gives 0t = and 0 2 2 2(40 0 m/s) 8 16 s. 9 80 m/s y y v t a .= − = − = + . − . (d) 0,yv = 0 40 0 m/s,yv = + . 29 80 m/s .ya = − . 0y y yv v a t= + gives 0 2 0 40 0 m/s 4 08 s. 9 80 m/s y y y v v t a − − .= = = . − . (e) The acceleration is 29 80 m/s ,. downward, at all points in the motion. (f) The graphs are sketched in Figure 2.48. EVALUATE: 0yv = at the maximum height. The time to reach the maximum height is half the total time in the air, so the answer in part (d) is half the answer in part (c). Also note that 2 04 s 4 08 s 6 12 s.. < . < . The boulder is going upward until it reaches its maximum height and after the maximum height it is traveling downward. Figure 2.48 2.49. IDENTIFY: The rock has a constant downward acceleration of 9.80 m/s2. The constant-acceleration kinematics formulas apply. SET UP: The formulas 210 0 2y yy y v t a t= + + and 2 2 0 02 ( )y y yv v a y y= + − both apply. Call +y upward. First find the initial velocity and then the final speed. EXECUTE: (a) 6.00 s after it is thrown, the rock is back at its original height, so y = y0 at that instant. Using ay = –9.80 m/s2 and t = 6.00 s, the equation 210 0 2y yy y v t a t= + + gives v0y = 29.4 m/s. When the rock 2-20 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EVALUATE: The acceleration is constant until 1 3 ms,t = . and then it is zero. 2980 cm/s .g = The acceleration during the first 1.3 ms is much larger than this and gravity can be neglected for the portion of the jump that we are considering. Figure 2.54 2.55. IDENTIFY: The sprinter’s acceleration is constant for the first 2.0 s but zero after that, so it is not constant over the entire race. We need to break up the race into segments. SET UP: When the acceleration is constant, the formula 00 2 x xv vx x t+⎛ ⎞− = ⎜ ⎟ ⎝ ⎠ applies. The average velocity is av- .x xv t Δ= Δ EXECUTE: (a) 00 0 10 0 m/s (2 0 s) 10 0 m. 2 2 x xv vx x t+ + .⎛ ⎞ ⎛ ⎞− = = . = .⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (b) (i) 40.0 m at 10.0 m/s so time at constant speed is 4.0 s. The total time is 6.0 s, so av- 50 0 m 8 33 m/s. 6 0 sx xv t Δ .= = = . Δ . (ii) He runs 90.0 m at 10.0 m/s so the time at constant speed is 9.0 s. The total time is 11.0 s, so av- 100 m 9 09 m/s. 11 0 sx v = = . . (iii) He runs 190 m at 10.0 m/s so time at constant speed is 19.0 s. His total time is 21.0 s, so av- 200 m 9 52 m/s. 21 0 sx v = = . . EVALUATE: His average velocity keeps increasing because he is running more and more of the race at his top speed. 2.56. IDENTIFY: We know the vertical position of the lander as a function of time and want to use this to find its velocity initially and just before it hits the lunar surface. SET UP: By definition, ( ) ,y dyv t dt = so we can find vy as a function of time and then evaluate it for the desired cases. EXECUTE: (a) ( ) 2 .y dyv t c dt dt = = − + At 0,t = ( ) 60 0 m/s.yv t c= − = − . The initial velocity is 60.0 m/s downward. (b) ( ) 0y t = says 2 0.b ct dt− + = The quadratic formula says 28 57 s 7 38 s.t = . ± . It reaches the surface at 21 19 s.t = . At this time, 260 0 m/s 2(1 05 m/s )(21 19 s) 15 5 m/s.yv = − . + . . = − . EVALUATE: The given formula for y(t) is of the form y = y0 + v0yt + 12 at 2. For part (a), v0y = −c = −60 m/s. 2.57. IDENTIFY: In time St the S-waves travel a distance S Sd v t= and in time Pt the P-waves travel a distance P P.d v t= SET UP: S P 33 st t= + EXECUTE: S P 1 133 s. 33 s and = 250 km. 3.5 km/s 6.5 km/s d d d d v v ⎛ ⎞ = + − =⎜ ⎟ ⎝ ⎠ Motion Along a Straight Line 2-21 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EVALUATE: The times of travel for each wave are S 71st = and P 38 s.t = 2.58. IDENTIFY: The brick has a constant downward acceleration, so we can use the usual kinematics formulas. We know that it falls 40.0 m in 1.00 s, but we do not know which second that is. We want to find out how far it falls in the next 1.00-s interval. SET UP: Let the +y direction be downward. The final velocity at the end of the first 1.00-s interval will be the initial velocity for the second 1.00-s interval. ay = 9.80 m/s2 and the formula 210 0 2y yy y v t a t=− + applies. EXECUTE: (a) First find the initial speed at the beginning of the first 1.00-s interval. Applying the above formula with ay = 9.80 m/s2, t = 1.00 s, and y – y0 = 40.0 m, we get v0y = 35.1 m/s. At the end of this 1.00-s interval, the velocity is vy = 35.1 m/s + (9.80 m/s2)(1.00 s) = 44.9 m/s. This is v0y for the next 1.00-s interval. Using 210 0 2y yy y v t a t=− + with this initial velocity gives y – y0 = 49.8 m. EVALUATE: The distance the brick falls during the second 1.00-s interval is greater than during the first 1.00-s interval, which it must be since the brick is accelerating downward. 2.59. IDENTIFY: The average velocity is av- .x xv t Δ= Δ SET UP: Let x+ be upward. EXECUTE: (a) av- 1000 m 63 m 197 m/s 4.75 sx v −= = (b) av- 1000 m 0 169 m/s 5.90 sx v −= = EVALUATE: For the first 1.15 s of the flight, av- 63 m 0 54.8 m/s. 1.15 sx v −= = When the velocity isn’t constant the average velocity depends on the time interval chosen. In this motion the velocity is increasing. 2.60. IDENTIFY: Use constant acceleration equations to find 0x x− for each segment of the motion. SET UP: Let x+ be the direction the train is traveling. EXECUTE: 0t = to 14.0 s: 2 2 21 10 0 2 2 (1.60 m/s )(14.0 s) 157 m.x xx x v t a t− = + = = At 14.0 s,t = the speed is 20 (1.60 m/s )(14.0 s) 22.4 m/s.x x xv v a t= + = = In the next 70.0 s, 0xa = and 0 0 (22.4 m/s)(70.0 s) 1568 m.xx x v t− = = = For the interval during which the train is slowing down, 0 22.4 m/s,xv = 23.50 m/sxa = − and 0.xv = 2 2 0 02 ( )x x xv v a x x= + − gives 2 2 2 0 0 2 0 (22.4 m/s) 72 m. 2 2( 3.50 m/s ) x x x v vx x a − −− = = = − The total distance traveled is 157 m 1568 m 72 m 1800 m.+ + = EVALUATE: The acceleration is not constant for the entire motion, but it does consist of constant acceleration segments, and we can use constant acceleration equations for each segment. 2.61. IDENTIFY: When the graph of xv versus t is a straight line the acceleration is constant, so this motion consists of two constant acceleration segments and the constant acceleration equations can be used for each segment. Since xv is always positive the motion is always in the x+ direction and the total distance moved equals the magnitude of the displacement. The acceleration xa is the slope of the xv versus t graph. SET UP: For the 0t = to 10 0 st = . segment, 0 4 00 m/sxv = . and 12 0 m/s.xv = . For the 10 0 st = . to 12 0 s. segment, 0 12 0 m/sxv = . and 0.xv = EXECUTE: (a) For 0t = to 10 0 s,t = . 00 4 00 m/s 12 0 m/s (10 0 s) 80 0 m. 2 2 x xv vx x t+ . + .⎛ ⎞ ⎛ ⎞− = = . = .⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ For 10 0 st = . to 12 0 s,t = . 0 12 0 m/s 0 (2 00 s) 12 0 m. 2 x x . +⎛ ⎞− = . = .⎜ ⎟ ⎝ ⎠ The total distance traveled is 92.0 m. (b) 0 80 0 m 12 0 m 92 0 mx x− = . + . = . 2-22 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (c) For 0t = to 10.0 s, 212 0 m/s 4 00 m/s 0 800 m/s . 10 0 sx a . − .= = . . For 10 0 st = . to 12.0 s, 20 12 0 m/s 6 00 m/s . 2 00 sx a − .= = − . . The graph of xa versus t is given in Figure 2.61. EVALUATE: When xv and xa are both positive, the speed increases. When xv is positive and xa is negative, the speed decreases. Figure 2.61 2.62. IDENTIFY: Apply 210 0 2x xx x v t a t− = + to the motion of each train. A collision means the front of the passenger train is at the same location as the caboose of the freight train at some common time. SET UP: Let P be the passenger train and F be the freight train. For the front of the passenger train 0 0x = and for the caboose of the freight train 0 200 m.x = For the freight train F 15 0 m/sv = . and F 0.a = For the passenger train P 25 0 m/sv = . and 2 P 0 100 m/s .a = − . EXECUTE: (a) 210 0 2x xx x v t a t− = + for each object gives 21 P P P2x v t a t= + and F F200 m .x v t= + Setting P Fx x= gives 21 P P F2 200 m .v t a t v t+ = + 2 2(0 0500 m/s ) (10 0 m/s) 200 m 0.t t. − . + = The quadratic formula gives ( )21 10 0 (10 0) 4(0 0500)(200) s (100 77 5) s.0 100t = + . ± . − . = ± .. The collision occurs at 100 s 77 5 s 22 5 s.t = − . = . The equations that specify a collision have a physical solution (real, positive t), so a collision does occur. (b) 2 21P 2(25 0 m/s)(22 5 s) ( 0 100 m/s )(22 5 s) 537 m.x = . . + − . . = The passenger train moves 537 m before the collision. The freight train moves (15 0 m/s)(22 5 s) 337 m.. . = (c) The graphs of Fx and Px versus t are sketched in Figure 2.62. EVALUATE: The second root for the equation for t, 177 5 st = . is the time the trains would meet again if they were on parallel tracks and continued their motion after the first meeting. Figure 2.62 2.63. IDENTIFY and SET UP: Apply constant acceleration kinematics equations. Find the velocity at the start of the second 5.0 s; this is the velocity at the end of the first 5.0 s. Then find 0x x− for the first 5.0 s. Motion Along a Straight Line 2-25 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 0 20 0 m/s,xv = . 20 600 m/s ,xa = . 15 86 s,t = . 0 ?x x− = 2 2 21 1 0 0 2 2(20 0 m/s)(15 86 s) (0 600 m/s )(15 86 s)x xx x v t a t− = + = . . + . . 0 317 2 m 75 5 m 393 mx x− = . + . = . (c) In coordinates fixed to the earth: 2 0 20 0 m/s (0 600 m/s )(15 86 s) 29 5 m/sx x xv v a t= + = . + . . = . EVALUATE: In 15.86 s the truck travels 0 (20 0 m/s)(15 86 s) 317 2 mx x− = . . = . . The car travels 392 7 m 317 2 m 75 m. − . = farther than the truck, which checks with part (a). In coordinates attached to the truck, for the car 0 0,xv = 9 5 m/sxv = . and in 15.86 s the car travels 00 75 m,2 x xv vx x t+⎛ ⎞− = =⎜ ⎟ ⎝ ⎠ which checks with part (a). 2.68. IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use ( ) xx dva t dt = and 0 0 ( ) . t xx x v t dt= + ∫ SET UP: 11 1 n nt dt t n += +∫ for 0.n ≥ EXECUTE: (a) 2 310 0 30( ) [ ] . t x t x t dt x t tα β α β= + − = + −∫ 0x = at 0t = gives 0 0x = and 3 3 31 3( ) (4 00 m/s) (0 667 m/s ) .x t t t t tα β= − = . − . 3( ) 2 (4 00 m/s ) .xx dva t t t dt β= = − = − . (b) The maximum positive x is when 0xv = and 0.xa < 0xv = gives 2 0tα β− = and 3 4 00 m/s 1 41 s. 2 00 m/s t α β .= = = . . At this t, xa is negative. For 1 41 s,t = . 3 3(4 00 m/s)(1 41 s) (0 667 m/s )(1 41 s) 3 77 m.x = . . − . . = . EVALUATE: After 1 41 st = . the object starts to move in the x− direction and goes to x = −∞ as .t → ∞ 2.69. (a) IDENTIFY and SET UP: Integrate ( )xa t to find ( )xv t and then integrate ( )xv t to find ( )x t . We know ( ) ,xa t tα β= + with 22 00 m/sα = − . and 33 00 m/sβ = . . EXECUTE: 210 0 0 20 0 ( ) t t x x x x xv v a dt v t dt v t tα β α β= + = + + = + +∫ ∫ 2 2 31 1 1 0 0 0 0 02 2 60 0 ( ) t t x x xx x v dt x v t t dt x v t t tα β α β= + = + + + = + + +∫ ∫ At 0,t = 0x x= . To have 0x x= at 1 4 00 st = . requires that 2 31 1 0 1 1 12 6 0xv t t tα β+ + = . Thus 2 3 2 21 1 1 10 1 16 2 6 2(3 00 m/s )(4 00 s) ( 2 00 m/s )(4 00 s) 4 00 m/sxv t tβ α= − − = − . . − − . . = − . . (b) With 0xv as calculated in part (a) and 4 00 s,t = . 2 2 3 21 1 0 2 24 00 m/s ( 2 00 m/s )(4 00 s) (3 00 m/s )(4 00 s) 12 0 m/sx xv v t tα β= + + = − . + − . . + . . = + . . EVALUATE: 0xa = at 0 67 st = . . For 0 67 s,t > . 0xa > . At 0,t = the particle is moving in the -directionx− and is speeding up. After 0 67 s,t = . when the acceleration is positive, the object slows down and then starts to move in the -directionx+ with increasing speed. 2.70. IDENTIFY: Find the distance the professor walks during the time t it takes the egg to fall to the height of his head. SET UP: Let y+ be downward. The egg has 0 0yv = and 29 80 m/s .ya = . At the height of the professor’s head, the egg has 0 44 2 m.y y− = . EXECUTE: 210 0 2y yy y v t a t− = + gives 0 2 2( ) 2(44 2 m) 3 00 s. 9 80 m/sy y yt a − .= = = . . The professor walks a distance 0 0 (1 20 m/s)(3 00 s) 3 60 m.xx x v t− = = . . = . Release the egg when your professor is 3.60 m from the point directly below you. 2-26 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. EVALUATE: Just before the egg lands its speed is 2(9 80 m/s )(3 00 s) 29 4 m/s.. . = . It is traveling much faster than the professor. 2.71. IDENTIFY: Use the constant acceleration equations to establish a relationship between maximum height and acceleration due to gravity and between time in the air and acceleration due to gravity. SET UP: Let y+ be upward. At the maximum height, 0.yv = When the rock returns to the surface, 0 0.y y− = EXECUTE: (a) 2 20 02 ( )y y yv v a y y= + − gives 21 02 ,y ya H v= − which is constant, so E E M M.a H a H= 2 E M E 2 M 9 80 m/s 2 64 . 3 71 m/s aH H H H a ⎛ ⎞ ⎛ ⎞.= = = .⎜ ⎟ ⎜ ⎟.⎝ ⎠⎝ ⎠ (b) 210 0 2y yy y v t a t− = + with 0 0y y− = gives 02 ,y ya t v= − which is constant, so E E M M.a T a T= E M E M 2 64 .aT T T a ⎡ ⎤ = = .⎢ ⎥ ⎣ ⎦ EVALUATE: On Mars, where the acceleration due to gravity is smaller, the rocks reach a greater height and are in the air for a longer time. 2.72. IDENTIFY: Calculate the time it takes her to run to the table and return. This is the time in the air for the thrown ball. The thrown ball is in free-fall after it is thrown. Assume air resistance can be neglected. SET UP: For the thrown ball, let y+ be upward. 29 80 m/s .ya = − . 0 0y y− = when the ball returns to its original position. The constant-acceleration kinematics formulas apply. EXECUTE: (a) It takes her 5 50 m 1.833 s 3.00 m/s . = to reach the table and an equal time to return, so the total time ball is in the air is 3.667 s. For the ball, 0 0,y y− = 3.667 st = and 29 80 m/s .ya = − . 21 0 0 2y yy y v t a t− = + gives 21 1 0 2 2 ( 9 80 m/s )(3.667 s) 18.0 m/s.y yv a t= − = − − . = (b) Find 0y y− when 1.833 s.t = 2 2 21 1 0 0 2 2(18.0 m/s)(1.833 s) ( 9 80 m/s )(1.833 s) 16.5 m.y yy y v t a t− = + = + − . = EVALUATE: It takes the ball the same amount of time to reach its maximum height as to return from its maximum height, so when she is at the table the ball is at its maximum height. Note that this large maximum height requires that the act either be done outdoors, or in a building with a very high ceiling. 2.73. (a) IDENTIFY: Consider the motion from when he applies the acceleration to when the shot leaves his hand. SET UP: Take positive y to be upward. 0 0,yv = ?,yv = 235 0 m/s ,ya = . 0 0 640 m,y y− = . 2 2 0 02 ( )y y yv v a y y= + − EXECUTE: 202 ( ) 2(35 0 m/s )(0 640 m) 6 69 m/sy yv a y y= − = . . = . (b) IDENTIFY: Consider the motion of the shot from the point where he releases it to its maximum height, where 0.v = Take 0y = at the ground. SET UP: 0 2 20 m,y = . ?,y = 29 80 m/sya = − . (free fall), 0 6 69 m/syv = . (from part (a), 0yv = at maximum height), 2 20 02 ( )y y yv v a y y= + − EXECUTE: 2 2 2 0 0 2 0 (6 69 m/s) 2 29 m, 2 2( 9 80 m/s ) y y y v v y y a − − .− = = = . − . 2 20 m 2 29 m 4 49 my = . + . = . . (c) IDENTIFY: Consider the motion of the shot from the point where he releases it to when it returns to the height of his head. Take 0y = at the ground. SET UP: 0 2 20 m,y = . 1 83 m,y = . 29 80 m/sya = − . , 0 6 69 m/s,yv = + . ?t = 21 0 0 2y yy y v t a t− = + EXECUTE: 2 2121 83 m 2 20 m (6 69 m/s) ( 9 80 m/s )t t. − . = . + − . 2 2(6 69 m/s) (4 90 m/s ) ,t t= . − . 24 90 6 69 0 37 0,t t. − . − . = with t in seconds. Use the quadratic formula to solve for t: Motion Along a Straight Line 2-27 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. ( )21 6 69 (6 69) 4(4 90)( 0 37) 0 6830 0 7362.9 80t = . ± . − . − . = . ± .. Since t must be positive, 0 6830 s 0 7362 s 1 42 s.t = . + . = . EVALUATE: Calculate the time to the maximum height: 0y y yv v a t,= + so 0( )/y y yt v v a= − = 2(6 69 m/s)/( 9 80 m/s ) 0 68 s− . − . = . . It also takes 0.68 s to return to 2.2 m above the ground, for a total time of 1.36 s. His head is a little lower than 2.20 m, so it is reasonable for the shot to reach the level of his head a little later than 1.36 s after being thrown; the answer of 1.42 s in part (c) makes sense. 2.74. IDENTIFY: The flowerpot is in free-fall. Apply the constant acceleration equations. Use the motion past the window to find the speed of the flowerpot as it reaches the top of the window. Then consider the motion from the windowsill to the top of the window. SET UP: Let y+ be downward. Throughout the motion 29 80 m/s .ya = + . The constant-acceleration kinematics formulas all apply. EXECUTE: Motion past the window: 0 1 90 m,y y− = . 0 380 s,t = . 29 80 m/s .ya = + . 21 0 0 2y yy y v t a t− = + gives 20 1 10 2 2 1 90 m (9 80 m/s )(0 380 s) 3.138 m/s. 0 380 sy y y yv a t t − .= − = − . . = . This is the velocity of the flowerpot when it is at the top of the window. Motion from the windowsill to the top of the window: 0 0,yv = 2 466 m/s,yv = . 29 80 m/s .ya = + . 2 2 0 02 ( )y y yv v a y y= + − gives 2 2 2 0 0 2 (3.138 m/s) 0 0 502 m. 2 2(9 80 m/s ) y y y v v y y a − −− = = = . . The top of the window is 0.502 m below the windowsill. EVALUATE: It takes the flowerpot 0 2 3.138 m/s 0 320 s 9 80 m/s y y y v v t a − = = = . . to fall from the sill to the top of the window. Our result says that from the windowsill the pot falls 0 502 m 1 90 m 2.4 m. + . = in 0 320 s 0 380 s 0 700 s.. + . = . 2 2 21 10 0 2 2 (9 80 m/s )(0 700 s) 2 4 m,y yy y v t a t− = + = . . = . which checks. 2.75. IDENTIFY: Two stones are thrown up with different speeds. (a) Knowing how soon the faster one returns to the ground, how long it will take the slow one to return? (b) Knowing how high the slower stone went, how high did the faster stone go? SET UP: Use subscripts f and s to refer to the faster and slower stones, respectively. Take y+ to be upward and 0 0y = for both stones. 0f 0s3v v= . When a stone reaches the ground, 0y = . The constant- acceleration formulas 210 0 2y yy y v t a t= + + and 2 2 0 02 ( )y y yv v a y y= + − both apply. EXECUTE: (a) 210 0 2y yy y v t a t= + + gives 02 y y v a t = − . Since both stones have the same ,ya 0f 0s f s v v t t = and ( )0s 1s f 3 0f (10 s) 3 3 svt t v ⎛ ⎞ = = = . .⎜ ⎟ ⎝ ⎠ (b) Since 0yv = at the maximum height, then 2 2 0 02 ( )y y yv v a y y= + − gives 2 0 2 y y v a y = − . Since both have the same ,ya 2 2 0f 0s f s v v y y = and 2 0f f s 0s 9vy y H v ⎛ ⎞ = = .⎜ ⎟ ⎝ ⎠ EVALUATE: The faster stone reaches a greater height so it travels a greater distance than the slower stone and takes more time to return to the ground. 2.76. IDENTIFY: The motion of the rocket can be broken into 3 stages, each of which has constant acceleration, so in each stage we can use the standard kinematics formulas for constant acceleration. But the acceleration is not the same throughout all 3 stages. SET UP: The formulas 00 ,2 y yv vy y t +⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ 2 0 0 1 , 2y y y y v t a t− = + and 0y y yv v a t= + apply. 2-30 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Putting in C = 0.625 m/s3 and t = 8.00 s gives an answer of 107 m. EVALUATE: The standard kinematics formulas are of no use in this problem since the acceleration varies with time. 2.82. IDENTIFY: Both objects are in free-fall and move with constant acceleration 29 80 m/s ,. downward. The two balls collide when they are at the same height at the same time. SET UP: Let y+ be upward, so 29 80 m/sya = − . for each ball. Let 0y = at the ground. Let ball A be the one thrown straight up and ball B be the one dropped from rest at height H. 0 0,Ay = 0 .By H= EXECUTE: (a) 210 0 2y yy y v t a t− = + applied to each ball gives 21 0 2Ay v t gt= − and 21 2 .By H gt= − A By y= gives 2 21 1 0 2 2v t gt H gt− = − and 0 .Ht v = (b) For ball A at its highest point, 0yAv = and 0y y yv v a t= + gives 0 . vt g = Setting this equal to the time in part (a) gives 0 0 H v v g = and 2 0 .vH g = EVALUATE: In part (a), using 0 Ht v = in the expressions for Ay and By gives 2 0 1 . 2A B gHy y H v ⎛ ⎞ = = −⎜ ⎟ ⎝ ⎠ H must be less than 2 02v g in order for the balls to collide before ball A returns to the ground. This is because it takes ball A time 02vt g = to return to the ground and ball B falls a distance 2 2 01 2 2vgt g = during this time. When 2 02vH g = the two balls collide just as ball A reaches the ground and for H greater than this ball A reaches the ground before they collide. 2.83. IDENTIFY and SET UP: Use /xv dx dt= and /x xa dv dt= to calculate ( )xv t and ( )xa t for each car. Use these equations to answer the questions about the motion. EXECUTE: 2 ,Ax t tα β= + 2 ,AAx dxv t dt α β= = + 2AxAx dva dt β= = 2 3,Bx t tγ δ= − 22 3 ,BBx dxv t t dt γ δ= = − 2 6BxBx dva t dt γ δ= = − (a) IDENTIFY and SET UP: The car that initially moves ahead is the one that has the larger 0xv . EXECUTE: At 0,t = Axv α= and 0Bxv = . So initially car A moves ahead. (b) IDENTIFY and SET UP: Cars at the same point implies A Bx x= . 2 2 3t t t tα β γ δ+ = − EXECUTE: One solution is 0,t = which says that they start from the same point. To find the other solutions, divide by t: 2t t tα β γ δ+ = − 2 ( ) 0t tδ β γ α+ − + = ( ) ( )2 21 1( ) ( ) 4 1 60 (1 60) 4(0 20)(2 60) 4 00 s 1 73 s2 0 40t β γ β γ δαδ= − − ± − − = + . ± . − . . = . ± .. So A Bx x= for 0,t = 2 27 st = . and 5 73 st = . . EVALUATE: Car A has constant, positive xa . Its xv is positive and increasing. Car B has 0 0xv = and xa that is initially positive but then becomes negative. Car B initially moves in the -directionx+ but then slows down and finally reverses direction. At 2 27 st = . car B has overtaken car A and then passes it. At 5 73 s,t = . car B is moving in the -directionx− as it passes car A again. Motion Along a Straight Line 2-31 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (c) IDENTIFY: The distance from A to B is B Ax x− . The rate of change of this distance is ( )B Ad x x dt − . If this distance is not changing, ( ) 0B Ad x x dt − = . But this says 0Bx Axv v− = . (The distance between A and B is neither decreasing nor increasing at the instant when they have the same velocity.) SET UP: Ax Bxv v= requires 22 2 3t t tα β γ δ+ = − EXECUTE: 23 2( ) 0t tδ β γ α+ − + = ( ) ( )2 21 12( ) 4( ) 12 3 20 4( 1 60) 12(0 20)(2 60)6 1 20t β γ β γ δαδ= − − ± − − = . ± − . − . .. 2 667 s 1 667 ,t s= . ± . so Ax Bxv v= for 1 00 st = . and 4 33 st = . . EVALUATE: At 1 00 s,t = . 5 00 m/sAx Bxv v= = . . At 4 33 s,t = . 13 0 m/sAx Bxv v= = . . Now car B is slowing down while A continues to speed up, so their velocities aren’t ever equal again. (d) IDENTIFY and SET UP: Ax Bxa a= requires 2 2 6 tβ γ δ= − EXECUTE: 2 2 3 2 80 m/s 1 20 m/s 2 67 s 3 3(0 20 m/s ) t γ β δ − . − .= = = . . . EVALUATE: At 0,t = ,Bx Axa a> but Bxa is decreasing while Axa is constant. They are equal at 2 67 st = . but for all times after that Bx Axa a< . 2.84. IDENTIFY: Interpret the data on a graph to draw conclusions about the motion of a glider having constant acceleration down a frictionless air track, starting from rest at the top. SET UP: The constant-acceleration kinematics formulas apply. Take the +x-axis along the surface of the track pointing downward. EXECUTE: (a) For constant acceleration starting from rest, we have 21 . 2 x x a t= Therefore a plot of x versus t2 should be a straight line, and the slope of that line should be ax/2. (b) To construct the graph of x versus t2, we can use readings from the graph given in the text to construct a table of values for x and t2, or we could use graphing software if available. The result is a graph similar to the one shown in Figure 2.84, which was obtained using software. A graph done by hand could vary slightly from this one, depending on how one reads the values on the graph in the text. The graph shown is clearly a straight line having slope 3.77 m/s2 and x-intercept 0.0092 m. Using the slope y-intercept form of the equation of a straight line, the equation of this line is x = 3.77t2 + 0.0092, where x is in meters and t in seconds. Figure 2.84 2-32 Chapter 2 © Copyright 2016 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (c) The slope of the straight line in the graph is ax/2, so ax = 2(3.77 m/s2) = 7.55 m/s2. (d) We know the distance traveled is 1.35 m, the acceleration is 7.55 m/s2, and the initial velocity is zero, so we use the equation 2 20 02 ( )x x xv v a x x= + − and solve for vx, giving vx = 4.51 m/s. EVALUATE: For constant acceleration in part (d), the average velocity is (4.51 m/s)/2 = 2.25 m/s. With this average velocity, the time for the glider to travel 1.35 m is x/vav = (1.35 m)/(2.25 m) = 0.6 s, which is approximately the value of t read from the graph in the text for x = 1.35 m. 2.85. IDENTIFY: A ball is dropped from rest and falls from various heights with constant acceleration. Interpret a graph of the square of its velocity just as it reaches the floor as a function of its release height. SET UP: Let y+ be downward since all motion is downward. The constant-acceleration kinematics formulas apply for the ball. EXECUTE: (a) The equation 2 20 02 ( )yy yv v a y y= + − applies to the falling ball. Solving for y – y0 and using v0y = 0 and ay = g, we get 2 0 .2 yvy y g − = A graph of y – y0 versus 2yv will be a straight line with slope 1/2g = 1/(19.6 m/s2) = 0.0510 s2/m. (b) With air resistance the acceleration is less than 9.80 m/s2, so the final speed will be smaller. (c) The graph will not be a straight line because the acceleration will vary with the speed of the ball. For a given release height, vy with air resistance is less than without it. Alternatively, with air resistance the ball will have to fall a greater distance to achieve a given velocity than without air resistance. The graph is sketched in Figure 2.85. Figure 2.85 EVALUATE: Graphing y – y0 versus 2yv for a set of data will tell us if the acceleration is constant. If the graph is a straight line, the acceleration is constant; if not, the acceleration is not constant. 2.86. IDENTIFY: Use data of acceleration and time for a model car to find information about its velocity and position. SET UP: From the table of data in the text, we can see that the acceleration is not constant, so the constant-acceleration kinematics formlas do not apply. Therefore we must use calculus. The equations 0 0 ( ) tx x xv t v a dt= + ∫ and 0 0( ) t xx t x v dt= + ∫ apply. EXECUTE: (a) Figure 2.86a shows the graph of ax versus t. From the graph, we find that the slope of the line is –0.5131 m/s3 and the a-intercept is 6.026 m/s2. Using the slope y-intercept equation of a straight line, the equation is a(t) = –0.513 m/s3 t + 6.026 m/s2, where t is in seconds and a is in m/s2.