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2-6Chapter 22.22. Set Up: aav, x Δυ xΔtSolve: (a) 0 s to 2 s: aav, x 0; 2 s to 4 s: aav, x 1.0 m/s 2 ; 4 s to 6 s: aav, x 1.5 m/s 2 ; 6 s to 8 s: aav, x 2.5 m/s 2 ;8 s to 10 s: aav, x 2.5 m/s 2 ; 10 s to 12 s: aav, x 2.5 m/s 2 ; 12 s to 14 s: aav, x 1.0 m/s 2 ; 14 s to 16 s: aav, x 0.The acceleration is not constant over the entire 16-s time interval. The acceleration is constant between 6 s and 12 s.(b) The graph of υ x versus t is given in the figure below. t 9 s: a x 2.5 m/s 2 ; t 13 s: a x 1.0 m/s 2 ; t 15 s: a x 0.Reflect: The acceleration is constant when the velocity changes at a constant rate. When the velocity is constant, theacceleration is zero.2.23. Set Up: 1 ft 0.3048 m, g 9.8 m/s 2 1 ft 2Solve: (a) 5 g 49 m/s 2 and 5 g (49 m/s 2 ) 160 ft/s 0.3048 m 1 ft 2(b) 60 g 590 m/s 2 and 60 g (590 m/s 2 ) 1900 ft/s03048m. 1g (c) (1.67 m/s 2 ) 0.17 g 9.8 m/s 2 () 1g (d) 24.3 m/s 2 2. 5 g 9.8 m/s 2 *2.24. Set Up: The acceleration ax equals the slope of the υ x versus t curve.Solve: The qualitative graphs of acceleration as a function of time are given in the figure below.The acceleration can be described as follows: (a) positive and constant, (b) positive and increasing, (c) negative andconstant, (d) positive and decreasing.Reflect: When υ x and a x have the same sign then the speed is increasing. In (c) the velocity and acceleration haveopposite signs and the speed is decreasing. 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.

Motion Along a Straight Line2-72.25. Set Up: The acceleration ax is the slope of the graph of υ x versus t.Solve: (a) Reading from the graph, at t 4.0 s, υ x 2.7 cm/s, to the right and at t 7.0 s, υ x 1.3 cm/s, to the left.(b) υ x versus t is a straight line with slope 8.0 cm/s 1.3 cm/s 2 . The acceleration is constant and equal to6.0 s1.3 cm/s 2 , to the left.(c) The graph of ax versus t is given in the figure below.*2.26. Set Up: Use ax Δυ xfor part (a) and υ x2 υ02x 2ax ( x x0 ) for part (b). Recall that a magnitude isΔtalways positive.Solve: (a) The initial speed is υ0 x 60 mi/h and the final speed is υ x 40 mi/h so the change in speed isΔυ x υ x υ0 x 40 mi/h 60 mi/h 20 mi/h. The time is Δt 3 s, which we will convert to hours. Thus, themagnitude of the acceleration isax Δυ x 20 mi/h 3600 s 42 1 h 2 10 mi/h3sΔt(b) The distance x x0 traveled while braking isυ x2 υ02x 2ax ( x x0 )( x x0 ) υ x2 υ02x2a x( 40 mi/h ) 2 ( 60 mi/h ) 2 5280 ft (2 2.4 104 mi/h 2) 1 mi 2 102 ftwhere we have retained a single significant digit because we are given the time of acceleration to a single significantdigit.Reflect: Note that we retained two significant digits for the acceleration in the intermediate step for part (b) andonly rounded down to a single significant digit at the end of the calculation.2.27. Set Up: 1 mph 0.4470 m/s and 1 m 3.281 ft. Let x0 0. υ0 x 0, t 2.0 s, and υ x 45 mph 20.1 m/s.υ υ0 x 20.1 m/s 0(a) υ x υ0 x axt and ax x 10 m/s 2t2.0 s 3.281 ft 2ax (10 m/s 2 ) 33 ft/s 1m (b) x x0 υ0 xt 1 2 1a xt 2 (10 m/s 2 )(2.0 s) 2 20 m, or x 12 (33 ft/s 2 )(2.0 s)2 66 ft2 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.

2-8Chapter 2*2.28. Set Up: Take the y direction to be upward. For part (a) we assume that the cat is in free fall with a y g.Since the cat falls a known distance, we can find its final velocity using υ y2 υ02y 2a y ( y y0 ). For parts (b) and (c)we assume that the cat has a constant (but unknown) acceleration due to its interaction with the floor. We may use theequations for constant acceleration.Solve: (a) Solving for υ y , we obtainυ y υ02y 2a y ( y y0 )Here we set Δy ( 4.0 ft)(1 m/3.28 ft) 1.22 m, a y g, and v y 0.υ y υ02y 2a y ( y y0 ) 2g( 1.22 m) 4.89 m/s 4.9 m/sWhere we choose the negative root since the cat is falling. The speed of the cat just before impact is the magnitude ofits velocity, which is 4.9 m/s.(b) During its impact with the floor, the cat is brought to rest over a distance of 12 cm. Thus, we have1υ0 y 4.89 m/s, υ y 0, and Δy 0.12 m. Solving Δy (υ y υ0 y )t for time, we obtain2t 2Δy2( 0.12 m) 0.049 s(υ y υ0 y ) 0 ( 4.89 m/s )(c) Solving υ y2 υ02y 2a y ( y y0 ) for a y , we obtain a y (υ y2 υ02y )2 Δy 02 ( 4.89 m/s) 2 99.6 m/s 2 . Since this2( 0.12 m)answer is only accurate to two significant figures, we can write it as 1.0 102 m/s 2 or approximately 10 g’s.Reflect: During free fall the cat has a negative velocity and a negative acceleration—so it is speeding up. In contrast,during impact with the ground the cat has a negative velocity and a positive acceleration—so it is slowing down.2.29. Set Up: Let x be in his direction of motion. Assume constant acceleration. (a) υ x 3(331 m/s) 993 m/s,υ0 x 0, and a x 5 g 49.0 m/s 2 ; (b) t 5.0 sυ υ0 x 993 m/s 0Solve: (a) υ x υ0 x a xt and t x 20.3 s2ax49.0 m/sYes, the time required is larger than 5.0 s.(b) υ x υ0 x a xt 0 (49.0 m/s 2 )(5.0 s) 245 m/s2.30. Set Up: Use υ x2 υ02x 2ax ( x x0 ) , where the distance is x x0 200 ft and the initial speed isυ0 x 60 mi/h for part (a) and υ0 x 100 mi/h for part (b). Recall that a magnitude is always positive.Solve: (a) Solving for the magnitude of the acceleration, inserting the given quantities, and converting miles to feet,we obtainυ x2 υ02x 2ax ( x x0 )ax υ x2 υ02x2 ( x x0 )( 0 mi/h ) 2 ( 60 mi/h ) 22 ( 200 ft ) 5280 ft 1 mi 4.8 103 mi/h 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.

Motion Along a Straight Line2-9(b) If the initial speed is 100 mi/h, the required acceleration isax υ x2 υ02x2 ( x x0 )( 0 mi/h ) 2 (100 mi/h ) 22 ( 200 ft ) 5280 ft 1 mi 1.3 105 mi/h 2Reflect: Note that the acceleration is negative because the speed of the car is decreasing. However, we are onlyinterested for this problem in the magnitude of the acceleration.2.31. Set Up: Let x be the direction the jet travels and take x0 0. ax 4 g 39.2 m/s 2 , υ x 4(331 m/s) 1324 m/s, and υ0 x 0.υ υ0 x 1324 m/s 0Solve: (a) υ x υ0 x a xt and t x 33.8 s2ax(b) x x0 υ0 xt 1 a t22 x 1 (39.2239.2 m/s2m/s )(33.8 s)2 2.24 104 m 22.4 km2.32. Set Up: Let x be the direction the person travels. υ x 0 (stops), t 36 ms 3.6 10 2 s, ax 60 g 588 m/s 2 . ax is negative since it is opposite to the direction of the motion.Solve: υ x υ0 x a xt so υ0x axt . Then x x0 υ0 xt 1 2axt gives x 12 axt 2 .21x ( 588 m/s 2 )(3.6 10 2 s) 2 38 cm2Reflect: We could also find the initial speed: υ0 x a xt ( 588 m/s 2 )(36 10 3 s) 21 m/s 47 mi/h2.33. Set Up: Take the x direction to be the direction of motion of the boulder.Solve: (a) Use the motion during the first second to find the acceleration. υ0 x 0, x0 0, x 2.00 m, andt 1.00 s.x x0 υ0 xt 1 22 x 2(2.00 m)axt and ax 2 4.00 m/s 222t(1.00 s)υ x υ0 x axt (4.00 m/s 2 )(1.00 s) 4.00 m/sFor the second second, υ0 x 4.00 m/s, a x 4.00 m/s 2 , and t 1.00 s.x x0 υ0 xt 1 2axt (4.00 m/s)(1.00 s) 12 (4.00 m/s 2 )(1.00 s) 2 6.00 m2We can also solve for the location at t 2.00 s, starting at t 0 :x x0 υ0 xt 12 axt 2 1(4.00 m/s 2 )(2.00 s) 2 8.00 m,2which agrees with 2.00 m in the first second and 6.00 m in the second second. The boulder speeds up so it travelsfarther in each successive second.(b) We have already found υ x 4.00 m/s after the first second. After the second second,υ x υ0 x a xt 4.00 m/s (4.00 m/s 2 )(1.00 s) 8.00 m/s 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.

2-10Chapter 2*2.34. Set Up: Let x be in the direction of motion of the bullet. υ0 x 0, x0 0, υ x 335 m/s, and x 0.127 m.Solve: (a) υ x2 υ02x 2a x ( x x0 ) andax (b) υ x υ0 x a xt so t υ x2 υ02x2( x x0 )υ x υ0 xax (335 m/s) 2 0 4.42 105 m/s 2 4.51 104 g2(0.127 m)335 m/s 04.42 105 m/s 2 0.758 ms υ υx Reflect: The acceleration is very large compared to g. In (b) we could also use ( x x0 ) 0 x t to calculate2 2( x x0 ) 2(0.127 m)t 0.758 ms.υx335 m/s2.35. Set Up: Take x in the direction to be in the direction in which the drag racer moves. Use1x x0 υ0 x t ax t 2 to find the acceleration with υ0 x 0, ( x x0 ) 0.25 mi, and t 10 s.21Solve: (a) x x0 υ0 x t ax t 2 gives2(b) The acceleration of gravity is 9.8 m/s2, which is about 20% greater than the acceleration of the drag racer.(c) To get the final speed, use υ x2 υ02x 2a x ( x x0 ) and insert the known quantities. This givesυ x2 υ02x 2ax ( x x0 )υ x υ02x 2ax ( x x0 ) 1609.34 m 0 2 ( 8.047 )( 0.25 mi ) 1 mi 80 m/sor about 180 mi/h.Solve: Note that we retained the positive sign in part (c) because the drag racer moves in the x direction. The resultfor part (c) seems like a reasonable speed for a drag racer.2.36. Set Up: 1 mi/h 1.466 ft/s. The car travels at constant speed during the reaction time. Let x be thedirection the car is traveling, so a x 12.0 ft/s 2 after the brakes are applied. 1.466 ft/s Solve: (a) υ0 x (15.0 mi/h) 22.0 ft/s. During the reaction time the car travels a distance of (22.0 ft/s) 1 mi/h (0.7 s) 15.4 ft.For the motion after the brakes are applied, υ0 x 22.0 ft/s, a x 12.0 ft/s 2 , and υ x 0. υ x2 υ02x 2a x ( x x0 )gives ( x x0 ) υ x2 υ02x0 (22.0 ft/s)2 20.2 ft.2( 12.0 ft/s 2 )The total distance is 15.4 ft 20.2 ft 35.6 ft.2a x 1.466 ft/s (b) υ0 x (55.0 mi/h) 80.6 ft/s. A calculation similar to that of part (a) gives a total stopping distance 1 mi/h of ( x x0 ) 56.4 ft 270.7 ft 327 ft. 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.

Motion Along a Straight Line2-11*2.37. Set Up: 0.250 mi 1320 ft. 60.0 mph 88.0 ft/s. Let x be the direction the car is traveling.Solve: (a) Braking: υ0 x 88.0 ft/s, x x0 146 ft, υ x 0. υ x2 υ02x 2a x ( x x0 ) givesax υ x2 υ02x2( x x0 ) 0 (88.0 ft/s)2 26.5 ft/s 22(146 ft)1 2axt gives2Speeding up: υ0 x 0, x x0 1320 ft, t 19.9 s. x x0 υ0 xt ax 2( x x0 ) t22(1320 ft)(19.9 s) 2 6.67 ft/s 2(b) υ x υ0 x a xt 0 (6.67 ft/s 2 )(19.9 s) 133 ft/s 90.5 mph(c) t υ x υ0 xax 0 88.0 ft/s 26.5 ft/s 2 3.32 sReflect: The magnitude of the acceleration while braking is much larger than when speeding up. That is why it takesmuch longer to go from 0 to 60 mph than to go from 60 to 0 mph.2.38. Set Up: Break this problem into two parts: In the first part, the car accelerates at ax 2 m/s 2 for anunknown distance until it reaches at speed of υ x 20 m/s. It accelerates over a time we’ll call t1, which we do notknow initially. In the second part of the problem, the car continues at this speed for an unknown time t2. We knowthat the total time t t1 t2 30 s.Solve: During the acceleration phase, use υ x2 υ0 x 2ax d1 to find the distance d1 that the car travels. This givesυ x2 υ0 x 2ax d1d1 υ x22a x( 20 m/s) 2(2 2 m/s 2) 100 mwhere we have used υ0 x 0. The time it takes to travel this distance may be found from11d1 υ0 x t1 ax t12 ax t1222t1 2d1 ax2 υ x2 ( 2ax ) ax υxax 20 m/s 10 s2 m/s 2Thus, the car travels for another 20 s at 20 m/s, covering an additional distance of 20 s 20 m/s 400 m. The totaldistance covered in the initial 30 s is thus 100 m 400 m 500 m.2.39. Set Up: A π r 2 and C 2π r , where r is the radius.Solve:A1r12 A2r2222 r 2r and A2 2 A1 1 A 4 Ar 1 r1 r 2r C1 C2 and C2 2 C1 1 C 2Cr1r2 r1 r1 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.

2-12Chapter 22.40. Set Up: Let L be the length of each side of the cube. The cube has 6 faces of area L2 , so A 6 L2 . V L3 .Solve:V1L13 A1L12V2L23 A2L2222 L 3L and A2 2 A1 1 A1 9 A1; surface area increases by a factor of 9. L1 L1 33 L 3L and V2 2 V1 1 V1 27V1; volume increases by a factor of 27. L1 L1 *2.41. Set Up: The volume of a cylinder of radius R and height H is given by V π R 2 H . We know the ratio ofthe heights of the two tanks and their volumes. From this information we can determine the ratio of their radii.Solve: We take the ratio of the volumes of the two tanks:have usedH largeH smallVlargeVsmall2 2 Rlarge 218 π Rlarge H large 1.20 , where we2150 π Rsmall H small Rsmall 1.20. Solving for the ratio of the radii we obtainRlargeRsmall 1 218 1.10. Thus, the larger 1.20 150 radius is 10% larger than the smaller radius.Reflect: All of the ratios used are dimensionless, and independent of the units used for measurement.42.42. Set Up: The formula for volume V π R 3 shows that V is proportional to R3.3Solve: If V R 3 then R V 1/3 . Thus, if V decreases by a factor of 8, R will decrease by a factor of 81/3 2.*2.43. Set Up: a A aB , x0 A x0 B 0, υ0 x, A υ0 x, B 0, and t A 2t B .Solve: (a) x x0 υ0 xt 1 211xxaxt gives x A a At A2 and xB aBt B 2 . a A aB gives A2 B2 and222tAtB22 t 1 xB B x A (250 km) 62.5 km 2 tA (b) υ x υ0 x a xt gives a A υAtAand aB υBtB. Since a A aB ,υAtA υBtBand tB 1 υ A (350 m/s) 175 m/s 2 tA υB Reflect: υ x is proportional to t and for υ0 x 0, x is proportional to t 2 .2.44. Set Up: a A 3aB and υ0 A υ0 B . Let x0 A x0 B 0. Since cars stop, υ A υ B 0. a Solve: (a) υ x2 υ02x 2ax ( x x0 ) gives a A x A aB xB , and xB A x A 3D aB a 1(b) υ x υ0 x axt gives a At A aBt B , so t A B t B T3a A 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.

Motion Along a Straight Line2-132.45. Set Up: We are given that a A 2aB and tB 2t A , where the subscripts refer to cyclist A and B.1 2at (where we have ignored their initial speed2because they start at rest). By taking the ratio of this equation for cyclist A and B, we findSolve: The motion of each cyclist must satisfy the equation d d A 12 a At A2 d B 12 aB tB2Using a A 2aB and tB 2t A , we get12a t 2d A 12 a At A21 1 2 BA 2 2d B 2 aB tB 12 aB ( 2t A )2We also know that υ 2 2ad for each cyclist, from which we obtain the ratioυ 2A 2a A d A υ B2 2aB d BUsing the previous result for the ratio of distances, we findυ 2A 2a A d A 2 ( 2aB ) 1 12aB 2υ B2 2aB d BυA 1υBReflect: We used the positive square root for the ratio of speeds because both cyclists move in the same direction.2.46. Set Up: Let y be upward. a y 9.80 m/s 2 . υ y 0 at the maximum height.Solve: (a) y y0 0.220 m, a y 9.80 m/s 2 , υ y 0. υ y2 υ02y 2a y ( y y0 ) givesυ0 y 2a y ( y y0 ) 2( 9.80 m/s 2 )(0.220 m) 2.08 m/s.(b) When the flea returns to ground, υ y υ0 y . υ y υ0 y a yt givest υ y υ0 yay 2.08 m/s 2.08 m/s 9.80 m/s 2 0.424 s(c) a 9.80 m/s 2 , downward, at all points in the motion.2.47. Set Up: Let y be downward. a y 9.80 m/s 2Solve: (a) υ0 y 0, t 2.50 s, a y 9.80 m/s 2 .y y0 υ0 yt 1 2 1a yt (9.80 m/s 2 )(2.50 s) 2 30.6 m22The building is 30.6 m tall.(b) υ y υ0 y a yt 0 (9.80 m/s 2 )(2.50 s) 24.5 m/s 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.

2-14Chapter 2(c) The graphs of a y , υ y , and y versus t are given in the figure below. Take y 0 at the ground.2.48. Set Up: Take downward to be the positive direction, so that the acceleration of gravity is g 9.8 m/s 2 . Theinitial speed of the marble is υ0 0 so we can ignore terms with initial speed. The distance traveled by the marble isx x0 830 m.11Solve: (a) We know x x0 at 2 gt 2 because the acceleration is that of gravity, so a g. Solving for the time22t givest 2 ( x x0 )g 2 ( 830 m )9.8 m/s 2 13 swhere we have retained only the positive square root as being physically relevant.(b) To find the speed υ of the marble just before it hits, we solve for υ inυ 2 2a ( x x0 ) 2 g ( x x0 )υ 2 g ( x x0 ) 2 ( 9.8 m/s 2 ) ( 830 m ) 1.3 102 m/swhere we have again retained only the positive square root as being physically relevant.*2.49. Set Up: Take y upward. υ y 0 at the maximum height. a y 0.379 g 3.71 m/s2 .Solve: Consider the motion from the maximum height back to the initial level. For this motion υ0 y 0 and1 2 1a yt ( 3.71 m/s 2 )(4.25 s) 2 33.5 m.22The ball went 33.5 m above its original position.t 4.25 s. y y0 υ0 yt (b) Consider the motion from just after it was hit to the maximum height. For this motion υ y 0 and t 4.25 s.υ y υ0 y a yt gives υ0 y a yt ( 3.71 m/s 2 )(4.25 s) 15.8 m/s.(c) The graphs are sketched in the figure below. 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.

Motion Along a Straight Line2-15Reflect: The answe

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