© Pearson Education Limited 2017. 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. click here to download 12–1. Starting from rest, a particle moving in a straight line has an acceleration of a = (2t - 6) m>s2, where t is in seconds. What is the particle’s velocity when t = 6 s, and what is its position when t = 11 s? Solution a = 2t - 6 dv = a dt L0 v dv = L0 t (2t - 6) dt v = t 2 - 6t ds = v dt L0 s ds = s = L0 t (t2 - 6t) dt t3 - 3t2 3 When t = 6 s, Ans. v = 0 When t = 11 s, Ans. s = 80.7 m @solutionmanual1 1 Ans: v = 0 s = 80.7 m © Pearson Education Limited 2017. 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. click here to download 12–2. The acceleration of a particle as it moves along a straight line is given by a = 14t 3 - 12 m>s2, where t is in seconds. If s = 2 m and v = 5 m>s when t = 0, determine the particle’s velocity and position when t = 5 s. Also, determine the total distance the particle travels during this time period. SOLUTION v L5 t dv = L0 3 (4 t - 1) dt v = t4 - t + 5 s L2 t ds = s = L0 (t4 - t + 5) dt 1 5 1 t - t2 + 5 t + 2 2 5 When t = 5 s, v = 625 m>s Ans. s = 639.5 m Ans. Since v Z 0 then d = 639.5 - 2 = 637.5 m Ans. Ans: v = 625 m>s s = 639.5 m d = 637.5 m @solutionmanual1 2 © Pearson Education Limited 2017. 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. click here to download 12–3. The velocity of a particle traveling in a straight line is given by v = (6t - 3t2) m>s, where t is in seconds. If s = 0 when t = 0, determine the particle’s deceleration and position when t = 3 s. How far has the particle traveled during the 3-s time interval, and what is its average speed? Solution v = 6t - 3t 2 a = dv = 6 - 6t dt At t = 3 s a = - 12 m>s2 Ans. ds = v dt L0 s ds = L0 t (6t - 3t2)dt s = 3t 2 - t 3 At t = 3 s Ans. s = 0 Since v = 0 = 6t - 3t 2, when t = 0 and t = 2 s. when t = 2 s, s = 3(2)2 - (2)3 = 4 m Ans. sT = 4 + 4 = 8 m ( vsp ) avg = sT 8 = = 2.67 m>s t 3 Ans. Ans: a = - 12 m>s2 s = 0 sT = 8 m (vsp)avg = 2.67 m>s @solutionmanual1 3 © Pearson Education Limited 2017. 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. click here to download *12–4. A particle is moving along a straight line such that its position is defined by s = (10t2 + 20) mm, where t is in seconds. Determine (a) the displacement of the particle during the time interval from t = 1 s to t = 5 s, (b) the average velocity of the particle during this time interval, and (c) the acceleration when t = 1 s. SOLUTION s = 10t2 + 20 (a) s|1 s = 10(1)2 + 20 = 30 mm s|5 s = 10(5)2 + 20 = 270 mm ¢s = 270 - 30 = 240 mm Ans. (b) ¢t = 5 - 1 = 4 s vavg = (c) a = 240 ¢s = = 60 mm>s ¢t 4 d2s = 20 mm s2 dt2 Ans. (for all t) Ans. Ans: ∆s = 240 mm vavg = 60 mm>s a = 20 mm>s2 @solutionmanual1 4 © Pearson Education Limited 2017. 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. click here to download 12–5. A particle moves along a straight line such that its position is defined by s = (t2 - 6t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s. Solution s = t2 - 6t + 5 v = ds = 2t - 6 dt a = dv = 2 dt v = 0 when t = 3 s t=0 = 5 s t = 3 = -4 s t=6 = 5 vavg = ∆s 0 = = 0 ∆t 6 ( vsp ) avg = Ans. sT 9 + 9 = = 3 m>s ∆t 6 Ans. a t = 6 = 2 m>s2 Ans. Ans: vavg = 0 (vsp)avg = 3 m>s @solutionmanual1 5 a t = 6 s = 2 m>s2 © Pearson Education Limited 2017. 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. click here to download 12–6. A stone A is dropped from rest down a well, and in 1 s another stone B is dropped from rest. Determine the distance between the stones another second later. SOLUTION + T s = s1 + v1 t + sA = 0 + 0 + 1 2 a t 2 c 1 (9.81)(2)2 2 sA = 19.62 m sA = 0 + 0 + 1 (9.81)(1)2 2 sB = 4.91 m ¢s = 19.62 - 4.91 = 14.71 m Ans. Ans: s = 14.71 m @solutionmanual1 6 © Pearson Education Limited 2017. 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. click here to download 12–7. A bus starts from rest with a constant acceleration of 1 m>s2. Determine the time required for it to attain a speed of 25 m>s and the distance traveled. SOLUTION Kinematics: v0 = 0, v = 25 m>s, s0 = 0, and ac = 1 m>s2. + B A: v = v0 + act 25 = 0 + (1)t t = 25 s + B A: Ans. v2 = v02 + 2ac(s - s0) 252 = 0 + 2(1)(s - 0) s = 312.5 m Ans. Ans: t = 25 s s = 312.5 m @solutionmanual1 7 © Pearson Education Limited 2017. 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. click here to download *12–8. A particle travels along a straight line with a velocity v = (12 - 3t 2) m>s, where t is in seconds. When t = 1 s, the particle is located 10 m to the left of the origin. Determine the acceleration when t = 4 s, the displacement from t = 0 to t = 10 s, and the distance the particle travels during this time period. SOLUTION v = 12 - 3t 2 (1) dv = - 6t dt a = s L-10 ds = L1 t=4 t v dt = = -24 m>s2 L1 t Ans. ( 12 - 3t 2 ) dt s + 10 = 12t - t 3 - 11 s = 12t - t 3 - 21 s t=0 s t = 10 = - 21 = -901 ∆s = -901 - ( -21) = -880 m Ans. From Eq. (1): v = 0 when t = 2s s t=2 = 12(2) - (2)3 - 21 = - 5 Ans. sT = (21 - 5) + (901 - 5) = 912 m Ans: a = - 24 m>s2 ∆s = - 880 m sT = 912 m @solutionmanual1 8 © Pearson Education Limited 2017. 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. click here to download 12–9. When two cars A and B are next to one another, they are traveling in the same direction with speeds vA and vB , respectively. If B maintains its constant speed, while A begins to decelerate at aA , determine the distance d between the cars at the instant A stops. A B d SOLUTION Motion of car A: v = v0 + act 0 = vA - aAt t = vA aA v2 = v20 + 2ac(s - s0) 0 = v2A + 2( - aA)(sA - 0) sA = v2A 2aA Motion of car B: sB = vBt = vB a vA vAvB b = aA aA The distance between cars A and B is sBA = |sB - sA| = ` v2A vAvB 2vAvB - v2A ` = ` ` aA 2aA 2aA Ans. Ans: @solutionmanual1 9 S BA = ` 2vA vB - v2A ` 2aA © Pearson Education Limited 2017. 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. click here to download 12–10. A particle travels along a straight-line path such that in 4 s it moves from an initial position sA = - 8 m to a position sB = +3 m. Then in another 5 s it moves from sB to sC = -6 m. Determine the particle’s average velocity and average speed during the 9-s time interval. SOLUTION Average Velocity: The displacement from A to C is ∆s = sC - SA = -6 - ( - 8) = 2 m. ∆s 2 vavg = = = 0.222 m>s Ans. ∆t 4 + 5 Average Speed: The distances traveled from A to B and B to C are sA S B = 8 + 3 = 11.0 m and sB S C = 3 + 6 = 9.00 m, respectively. Then, the total distance traveled is sTot = sA S B + sB S C = 11.0 + 9.00 = 20.0 m. (vsp)avg = sTot 20.0 = = 2.22 m>s ∆t 4 + 5 Ans. Ans: vavg = 0.222 m>s (vsp)avg = 2.22 m>s @solutionmanual1 10 © Pearson Education Limited 2017. 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. 12–11. click here to download Traveling with an initial speed of 70 km>h, a car accelerates at 6000 km>h2 along a straight road. How long will it take to reach a speed of 120 km>h? Also, through what distance does the car travel during this time? SOLUTION v = v1 + ac t 120 = 70 + 6000(t) t = 8.33(10 - 3) hr = 30 s Ans. v2 = v21 + 2 ac(s - s1) (120)2 = 702 + 2(6000)(s - 0) s = 0.792 km = 792 m Ans. Ans: t = 30 s s = 792 m @solutionmanual1 11 © Pearson Education Limited 2017. 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. click here to download *12–12. A particle moves along a straight line with an acceleration of a = 5>(3s1>3 + s 5>2) m>s2, where s is in meters. Determine the particle’s velocity when s = 2 m, if it starts from rest when s = 1 m . Use a numerical method to evaluate the integral. SOLUTION a = 5 1 3 5 A 3s + s2 B a ds = v dv 2 v 5 ds 1 3 L1 A 3s + s 0.8351 = 5 2 B = L0 v dv 1 2 v 2 v = 1.29 m>s Ans. Ans: v = 1.29 m>s @solutionmanual1 12 © Pearson Education Limited 2017. 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. click here to download 12–13. The acceleration of a particle as it moves along a straight line is given by a = (2t - 1) m>s2, where t is in seconds. If s = 1 m and v = 2 m>s when t = 0, determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during this time period. Solution a = 2t - 1 dv = a dt L2 v t L0 dv = (2t - 1)dt v = t2 - t + 2 dx = v dt Lt s ds = s = L0 t (t2 - t + 2)dt 1 3 1 t - t 2 + 2t + 1 3 2 When t = 6 s v = 32 m>s Ans. s = 67 m Ans. Since v ≠ 0 for 0 … t … 6 s, then Ans. d = 67 - 1 = 66 m Ans: v = 32 m>s s = 67 m d = 66 m @solutionmanual1 13 © Pearson Education Limited 2017. 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. click here to download 12–14. A train starts from rest at station A and accelerates at 0.5 m>s2 for 60 s. Afterwards it travels with a constant velocity for 15 min. It then decelerates at 1 m>s2 until it is brought to rest at station B. Determine the distance between the stations. SOLUTION Kinematics: For stage (1) motion, v0 = 0, s0 = 0, t = 60 s, and ac = 0.5 m>s2. Thus, + B A: s = s0 + v0t + s1 = 0 + 0 + + B A: 1 2 at 2 c 1 (0.5)(602) = 900 m 2 v = v0 + act v1 = 0 + 0.5(60) = 30 m>s For stage (2) motion, v0 = 30 m>s, s0 = 900 m, ac = 0 and t = 15(60) = 900 s. Thus, + B A: s = s0 + v0t + 1 2 at 2 c s2 = 900 + 30(900) + 0 = 27 900 m For stage (3) motion, v0 = 30 m>s, v = 0, s0 = 27 900 m and ac = - 1 m>s2. Thus, + B A: v = v0 + act 0 = 30 + ( - 1)t t = 30 s + : s = s0 + v0t + 1 2 at 2 c s3 = 27 900 + 30(30) + 1 (- 1)(302) 2 = 28 350 m = 28.4 km Ans. Ans: s = 28.4 km @solutionmanual1 14 © Pearson Education Limited 2017. 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. click here to download 12–15. A particle is moving along a straight line such that its velocity is defined as v = ( - 4s2) m>s, where s is in meters. If s = 2 m when t = 0, determine the velocity and acceleration as functions of time. SOLUTION v = - 4s2 ds = - 4s2 dt s L2 s - 2 ds = t L0 - 4 dt - s - 1| s2 = - 4t|t0 t = 1 -1 (s - 0.5) 4 s = 2 8t + 1 v = -4a a = 2 2 16 b = m>s 8t + 1 (8t + 1)2 Ans. 16(2)(8t + 1)(8) dv 256 = = m>s2 dt (8t + 1)4 (8t + 1)3 Ans. Ans: 16 m>s (8t + 1)2 256 a= m>s2 (8t + 1)3 v = @solutionmanual1 15 © Pearson Education Limited 2017. 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. click here to download *12–16. Determine the time required for a car to travel 1 km along a road if the car starts from rest, reaches a maximum speed at some intermediate point, and then stops at the end of the road. The car can accelerate at 1.5 m>s2 and decelerate at 2 m>s2. SOLUTION Using formulas of constant acceleration: v2 = 1.5 t1 x = 1 (1.5)(t21) 2 0 = v2 - 2 t2 1 (2)(t22) 2 1000 - x = v2t2 - Combining equations: t1 = 1.33 t2; v2 = 2 t2 x = 1.33 t22 1000 - 1.33 t22 = 2 t22 - t22 t2 = 20.702 s; t1 = 27.603 s t = t1 + t2 = 48.3 s Ans. Ans: t = 48.3 s @solutionmanual1 16 © Pearson Education Limited 2017. 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. click here to download 12–17. A particle is moving with a velocity of v0 when s = 0 and t = 0. If it is subjected to a deceleration of a = -kv3, where k is a constant, determine its velocity and position as functions of time. SOLUTION dn = - kn3 dt a = t n n - 3 dn = Ln0 L0 - k dt 1 -2 1n - n0- 22 = - kt 2 - n = a 2kt + a 1 -2 1 b b n20 Ans. ds = n dt s L0 s = s = t ds = L0 dt a2kt + a 2a 2kt + a 2k 1 2 1 bb v20 1 t 2 1 b b 3 n20 0 1 1 1 B ¢ 2kt + ¢ 2 ≤ ≤ - R n0 k n0 1 2 Ans. Ans: v = a2kt + s= @solutionmanual1 17 1 - 1>2 b v20 1 1 1>2 1 c a2kt + 2 b d k v0 v0 © Pearson Education Limited 2017. 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. click here to download 12–18. A particle is moving along a straight line with an initial velocity of 6 m>s when it is subjected to a deceleration of a = (- 1.5v1>2) m>s2, where v is in m>s. Determine how far it travels before it stops. How much time does this take? SOLUTION Distance Traveled: The distance traveled by the particle can be determined by applying Eq. 12–3. ds = vdv a s L0 v ds = v 1 L6 m>s - 1.5v2 v s = dv 1 L6 m>s - 0.6667 v2 dv 3 = a -0.4444v2 + 6.532 b m When v = 0, 3 s = - 0.4444a 0 2 b + 6.532 = 6.53 m Ans. Time: The time required for the particle to stop can be determined by applying Eq. 12–2. dt = dv a t L0 v dt = 1 t = - 1.333av2 b When v = 0, v 6 m>s dv 1 L6 m>s 1.5v 2 1 = a3.266 - 1.333v 2 b s 1 t = 3.266 - 1.333 a 0 2 b = 3.27 s Ans. Ans: s = 6.53 m t = 3.27 s @solutionmanual1 18 © Pearson Education Limited 2017. 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. click here to download 12–19. The acceleration of a rocket traveling upward is given by a = (6 + 0.02s) m>s2, where s is in meters. Determine the rocket’s velocity when s = 2 km and the time needed to reach this attitude. Initially, v = 0 and s = 0 when t = 0. SOLUTION b 6 m>s2 c ap b c sp vp ´ µ ¶ vp vp dvp 0 ´ µ ¶ sp1 2000 m dvp s dsp sp b c sp dsp 0 2 vp b sp 2 dsp vp t 0.02 s- 2 dt µ́ µ µ ¶ sp 0 c 2 sp 2 2 2b sp c sp 1 2 2b sp c sp dsp 2 2b sp1 c sp1 vp1 t1 µ́ µ µ ¶ sp1 vp1 1 2 dsp t1 322.49 m>s Ans. 19.27 s Ans. 2b sp c sp 0 Ans: vp1 = 322.49 t1 = 19.27 s @solutionmanual1 19 m s © Pearson Education Limited 2017. 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. click here to download *12–20. The acceleration of a rocket traveling upward is given by a = 16 + 0.02s2 m>s2, where s is in meters. Determine the time needed for the rocket to reach an altitude of s = 100 m. Initially, v = 0 and s = 0 when t = 0. SOLUTION a ds = n dv s s L0 16 + 0.02 s2 ds = 6 s + 0.01 s2 = n L0 n dn 1 2 n 2 n = 212 s + 0.02 s2 ds = n dt 100 L0 ds 212 s + 0.02 s2 1 20.02 t = L0 dt 1n B 212s + 0.02s2 + s20.02 + t = 5.62 s 12 2 20.02 R 100 = t 0 Ans. Ans: t = 5.62 s @solutionmanual1 20 © Pearson Education Limited 2017. 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. click here to download 12–21. When a train is traveling along a straight track at 2 m/s, it begins to accelerate at a = 160 v-42 m>s2, where v is in m/s. Determine its velocity v and the position 3 s after the acceleration. v s SOLUTION a = dv dt dt = dv a v 3 dt = L0 3 = dv -4 L2 60v 1 (v5 - 32) 300 v = 3.925 m>s = 3.93 m>s Ans. ads = vdv ds = s L0 ds = s = 1 5 vdv = v dv a 60 1 60 L2 3.925 v5 dv 1 v6 3.925 a b` 60 6 2 = 9.98 m Ans. Ans: v = 3.93 m>s s = 9.98 m @solutionmanual1 21 © Pearson Education Limited 2017. 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. click here to download 12–22. The acceleration of a particle along a straight line is defined by a = 12t - 92 m>s2, where t is in seconds. At t = 0, s = 1 m and v = 10 m>s. When t = 9 s, determine (a) the particle’s position, (b) the total distance traveled, and (c) the velocity. SOLUTION a = 2t - 9 v L10 t dv = L0 12t - 92 dt v - 10 = t2 - 9 t v = t2 - 9 t + 10 s L1 t ds = s-1 = s = L0 1t2 - 9t + 102 dt 13 t - 4.5 t2 + 10 t 3 13 t - 4.5 t2 + 10 t + 1 3 Note when v = t2 - 9 t + 10 = 0: t = 1.298 s and t = 7.701 s When t = 1.298 s, s = 7.13 m When t = 7.701 s, s = - 36.63 m When t = 9 s, s = -30.50 m (a) s = - 30.5 m (b) sTo t = (7.13 - 1) + 7.13 + 36.63 + (36.63 - 30.50) (c) Ans. sTo t = 56.0 m Ans. v = 10 m>s Ans. Ans: (a) s = - 30.5 m (b) sTot = 56.0 m (c) v = 10 m>s @solutionmanual1 22 © Pearson Education Limited 2017. 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. click here to download 12–23. If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation a = 9.81[1 - v2(10-4)] m>s2, where v is in m>s and the positive direction is downward. If the body is released from rest at a very high altitude, determine (a) the velocity when t = 5 s, and (b) the body’s terminal or maximum attainable velocity (as t : q ). SOLUTION Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2. (+ T ) dt = t L0 dv a v dv 2 L0 9.81[1 - (0.01v) ] dt = v t = v 1 dv dv c + d 9.81 L0 2(1 + 0.01v) L0 2(1 - 0.01v) 9.81t = 50ln a v = 1 + 0.01v b 1 - 0.01v 100(e0.1962t - 1) (1) e0.1962t + 1 a) When t = 5 s, then, from Eq. (1) v = b) If t : q , e0.1962t - 1 e0.1962t + 1 100[e0.1962(5) - 1] e0.1962(5) + 1 = 45.5 m>s Ans. : 1. Then, from Eq. (1) vmax = 100 m>s Ans. Ans: (a) v = 45.5 m>s (b) v max = 100 m>s @solutionmanual1 23 © Pearson Education Limited 2017. 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. click here to download *12–24. A sandbag is dropped from a balloon which is ascending vertically at a constant speed of 6 m>s. If the bag is released with the same upward velocity of 6 m>s when t = 0 and hits the ground when t = 8 s, determine the speed of the bag as it hits the ground and the altitude of the balloon at this instant. SOLUTION (+ T ) s = s0 + v0 t + 1 a t2 2 c h = 0 + ( -6)(8) + 1 (9.81)(8)2 2 = 265.92 m During t = 8 s, the balloon rises h¿ = vt = 6(8) = 48 m Altitude = h + h¿ = 265.92 + 48 = 314 m (+ T) Ans. v = v0 + ac t v = - 6 + 9.81(8) = 72.5 m s Ans. Ans: h = 314 m v = 72.5 m>s @solutionmanual1 24 © Pearson Education Limited 2017. 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. click here to download 12–25. A particle is moving along a straight line such that its acceleration is defined as a = (-2v) m>s2, where v is in meters per second. If v = 20 m>s when s = 0 and t = 0, determine the particle’s position, velocity, and acceleration as functions of time. Solution a = - 2v dv = - 2v dt v t dv -2 dt = L20 v L0 ln v = -2t 20 v = ( 20e -2t ) m>s a = L0 dv = dt Ans. ( - 40e -2t ) m>s2 s ds = v dt = L0 Ans. t (20e-2t)dt s = - 10e -2t t0 = - 10 ( e -2t - 1 ) s = 10 ( 1 - e -2t ) m Ans. Ans: v = ( 20e -2t ) m>s a = ( - 40e -2t ) m>s2 s = 10 ( 1 - e -2t ) m @solutionmanual1 25 © Pearson Education Limited 2017. 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. click here to download 12–26. The acceleration of a particle traveling along a straight line 1 1>2 is a = s m> s2, where s is in meters. If v = 0, s = 1 m 4 when t = 0, determine the particle’s velocity at s = 2 m. SOLUTION Velocity: + B A: v dv = a ds v L0 s v dv = v 1 1>2 s ds L1 4 s v2 2 = 1 s3>2 ` 2 0 6 1 v = 1 23 1s3>2 - 121>2 m>s When s = 2 m, v = 0.781 m>s. Ans. Ans: v = 0.781 m>s @solutionmanual1 26 © Pearson Education Limited 2017. 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. click here to download 12–27. When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf. If this variation of the acceleration can be expressed as a = 1g>v2f21v2f - v22, determine the time needed for the velocity to become v = vf>2 . Initially the particle falls from rest. SOLUTION g dv = a = ¢ 2 ≤ A v2f - v2 B dt vf v dy L0 v2f 2¿ - v = t g v2f L0 dt vf + v y g 1 ln ¢ ≤` = 2t 2vf vf - v 0 vf t = t = vf 2g vf 2g ln ¢ ln ¢ t = 0.549 a vf + v vf - v ≤ vf + vf> 2 vf - vf> 2 vf g ≤ b Ans. Ans: @solutionmanual1 27 t = 0.549 a vf g b © Pearson Education Limited 2017. 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. click here to download *12–28. A sphere is fired downwards into a medium with an initial speed of 27 m>s. If it experiences a deceleration of a = ( -6t) m>s2, where t is in seconds, determine the distance traveled before it stops. SOLUTION Velocity: v0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have A+TB dv = adt v L27 t dv = L0 -6tdt v = A 27 - 3t2 B m>s (1) At v = 0, from Eq. (1) 0 = 27 - 3t2 t = 3.00 s Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result v = 27 - 3t2 and applying Eq. 12–1, we have A+TB ds = vdt s L0 t ds = L0 A 27 - 3t2 B dt s = A 27t - t3 B m (2) At t = 3.00 s, from Eq. (2) s = 27(3.00) - 3.003 = 54.0 m Ans. Ans: s = 54.0 m @solutionmanual1 28 © Pearson Education Limited 2017. 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. click here to download 12–29. A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same instant another ball B is thrown upward from the ground with an initial velocity of 20 m>s. Determine the height from the ground and the time at which they pass. Solution Origin at roof: Ball A: 1 2 ( + c ) s = s0 + v0t + act 2 - s = 0 + 5t - 1 (9.81)t 2 2 Ball B: 1 2 ( + c ) s = s0 + v0t + act 2 - s = -30 + 20t - 1 (9.81)t 2 2 Solving, Ans. t = 2 s s = 9.62 m Distance from ground, Ans. d = (30 - 9.62) = 20.4 m Also, origin at ground, s = s0 + v0t + 1 2 at 2 c sA = 30 + 5t + 1 ( -9.81)t 2 2 sB = 0 + 20t + 1 ( - 9.81)t 2 2 Require sA = sB 30 + 5t + 1 1 ( -9.81)t 2 = 20t + ( -9.81)t 2 2 2 t = 2 s Ans. sB = 20.4 m Ans. @solutionmanual1 29 Ans: h = 20.4 m t = 2s © Pearson Education Limited 2017. 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. click here to download 12–30. A boy throws a ball straight up from the top of a 12-m high tower. If the ball falls past him 0.75 s later, determine the velocity at which it was thrown, the velocity of the ball when it strikes the ground, and the time of flight. SOLUTION Kinematics: When the ball passes the boy, the displacement of the ball in equal to zero. Thus, s = 0. Also, s0 = 0, v0 = v1, t = 0.75 s, and ac = - 9.81 m>s2. A+cB s = s0 + v0t + 1 2 at 2 c 0 = 0 + v110.752 + 1 1-9.81210.7522 2 v1 = 3.679 m>s = 3.68 m>s Ans. When the ball strikes the ground, its displacement from the roof top is s = - 12 m. Also, v0 = v1 = 3.679 m>s, t = t2, v = v2, and ac = - 9.81 m>s2. A+cB s = s0 + v0t + 1 2 at 2 c - 12 = 0 + 3.679t2 + 1 1- 9.812t22 2 4.905t22 - 3.679t2 - 12 = 0 t2 = 3.679 ; 21 - 3.67922 - 414.90521 -122 214.9052 Choosing the positive root, we have t2 = 1.983 s = 1.98 s Ans. Using this result, A+cB v = v0 + act v2 = 3.679 + 1 -9.81211.9832 = - 15.8 m>s = 15.8 m>s T Ans. @solutionmanual1 30 Ans: v1 = 3.68 m>s t2 = 1.98 s v2 = 15.8 m>s T