91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 1 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. •12–1. A car starts from rest and with constant acceleration achieves a velocity of 15 m>s when it travels a distance of 200 m. Determine the acceleration of the car and the time required. Kinematics: v0 = 0, v = 15 m>s, s0 = 0, and s = 200 m. A:B v2 = v0 2 + 2ac(s – s0) 152 = 02 + 2ac(200 – 0) ac = 0. 5625 m>s2 + A:B v = v0 + act 15 = 0 + 0. 5625t t = 26. 7 s Ans. Ans. 12–2. A train starts from rest at a station and travels with a constant acceleration of 1 m>s2. Determine the velocity of the train when t = 30 s and the distance traveled during this time. Kinematics: ac = 1 m>s2, v0 = 0, s0 = 0, and t = 30 s. + A:B + A:B v = v0 + act = 0 + 1(30) = 30 m>s s = s0 + v0t + 1 2 at 2 c Ans. = 0 + 0 + = 450 m 1 (1) A 302 B 2 Ans. 1 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 2 © 2010 Pearson Education, Inc. Upper Saddle River, NJ. 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. 12–3. An elevator descends from rest with an acceleration of 5 ft>s2 until it achieves a velocity of 15 ft>s. Determine the time required and the distance traveled. Kinematics: ac = 5 ft>s2, v0 = 0, v = 15 ft>s, and s0 = 0. A+TB v = v0 + act 15 = 0 + 5t t = 3s Ans. A+TB v2 = v0 2 + 2ac(s – s0) 152 = 02 + 2(5)(s – 0) s = 22. 5 ft Ans. 12–4. A car is traveling at 15 m>s, when the traffic light 50 m ahead turns yellow. Determine the required constant deceleration of the car and the time needed to stop the car at the light. Kinematics: v0 = 0, s0 = 0, s = 50 m and v0 = 15 m>s. + A:B v2 = v0 2 + 2ac(s – s0) 0 = 152 + 2ac(50 – 0) ac = -2. 25 m>s2 = 2. 25 m>s2 ; + A:B v = v0 + act 0 = 15 + (-2. 25)t t = 6. 67 s Ans. Ans. 2 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 3 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. •12–5. A particle is moving along a straight line with the acceleration a = (12t – 3t1/2) ft>s2, where t is in seconds. Determine the velocity and the position of the particle as a function of time. When t = 0, v = 0 and s = 15 ft. Velocity: + A:B dv = a dt v 0 L v t 0 L dv = v 0 = A 6t2 – 2t3>2 B 2 + A:B A 12t – 3t1>2 B dt t 0 v = A 6t2 – 2t3>2 B ft>s Ans. Position: Using this result and the initial condition s = 15 ft at t = 0 s, ds = v dt s 15 L ft s t ds = 0 L A 6t2 – 2t3>2 B dt 5>2 2 t t b 5 0 s 15 ft = a 2t3 s = a 2t3 – 4 5>2 t + 15b ft 5 Ans. 12–6. A ball is released from the bottom of an elevator which is traveling upward with a velocity of 6 ft>s. If the ball strikes the bottom of the elevator shaft in 3 s, determine the height of the elevator from the bottom of the shaft at the instant the ball is released. Also, find the velocity of the ball when it strikes the bottom of the shaft. Kinematics: When the ball is released, its velocity will be the same as the elevator at the instant of release. Thus, v0 = 6 ft>s. Also, t = 3 s, s0 = 0, s = -h, and ac = -32. 2 ft>s2. A+cB = s0 + v0t + 1 a t2 2 c 1 (-32. 2) A 32 B 2 Ans. -h = 0 + 6(3) + h = 127 ft A+cB v = v0 + act v = 6 + (-32. 2)(3) = -90. 6 ft>s = 90. 6 ft>s T Ans. 3 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 4 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–7. A car has an initial speed of 25 m>s and a constant deceleration of 3 m>s2. Determine the velocity of the car when t = 4 s.

What is the displacement of the car during the 4-s time interval? How much time is needed to stop the car? v = v0 + act v = 25 + (-3)(4) = 13 m>s ? s = s – s0 = v0 t + 1 a t2 2 c 1 (-3)(4)2 = 76 m 2 Ans. Ans. ?s = s – 0 = 25(4) + v = v0 + ac t 0 = 25 + (-3)(t) t = 8. 33 s Ans. *12–8. If a particle has an initial velocity of v0 = 12 ft>s to the right, at s0 = 0, determine its position when t = 10 s, if a = 2 ft>s2 to the left. + A:B 1 a t2 2 c 1 ( -2)(10)2 2 Ans. s = s0 + v0 t + = 0 + 12(10) + = 20 ft •12–9. The acceleration of a particle traveling along a straight line is a = k>v, where k is a constant.

If s = 0, v = v0 when t = 0, determine the velocity of the particle as a function of time t. Velocity: + A:B dt = t 0 L t 0 L dv a dv k>v v L0 1 vdv k v L0 v v dt = v = 22kt + v0 2 t = t 1 2 2v t2 = v 2k v0 0 dt = 1 A v2 – v0 2 B 2k Ans. 4 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 5 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–10.

Car A starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft>s2 until it reaches a speed of 80 ft>s. Afterwards it maintains this speed. Also, when t = 0, car B located 6000 ft down the road is traveling towards A at a constant speed of 60 ft>s. Determine the distance traveled by car A when they pass each other. 60 ft/s A B 6000 ft Distance Traveled: Time for car A to achives y = 80 ft>s can be obtained by applying Eq. 12–4. + A:B y = y0 + ac t 80 = 0 + 6t t = 13. 33 s The distance car A travels for this part of motion can be determined by applying Eq. 12–6. A:B y2 = y2 + 2ac (s – s0) 0 802 = 0 + 2(6)(s1 – 0) s1 = 533. 33 ft For the second part of motion, car A travels with a constant velocity of y = 80 ft>s and the distance traveled in t? = (t1 – 13. 33) s (t1 is the total time) is + A:B s2 = yt? = 80(t1 – 13. 33) Car B travels in the opposite direction with a constant velocity of y = 60 ft>s and the distance traveled in t1 is + A:B It is required that s1 + s2 + s3 = 6000 533. 33 + 80(t1 – 13. 33) + 60t1 = 6000 t1 = 46. 67 s The distance traveled by car A is sA = s1 + s2 = 533. 33 + 80(46. 67 – 13. 33) = 3200 ft Ans. s3 = yt1 = 60t1 5 91962_01_s12-p0001-0176 6/8/09 8:05 AM

Page 6 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. A particle travels along a straight line with a velocity v = (12 – 3t2) 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. v = 12 – 3t2 a = s -10 L s = dv = -6t 2 = -24 m>s2 dt t=4 t t 1 L (1) Ans. v dt = 1 L A 12 – 3t2 B dt s + 10 = 12t – t3 – 11 s = 12t – t3 – 21 s| t = 0 = -21 s|t = 10 = -901 ? s = -901 – ( -21) = -880 m From Eq. (1): v = 0 when t = 2s s|t = 2 = 12(2) – (2)3 – 21 = -5 sT = (21 – 5) + (901 – 5) = 912 m Ans. Ans. 6 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 7 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. *12–12.

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. Velocity: y0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have A+TB y 27 L dy = adt t 0 L dy = -6tdt [1] y = A 27 – 3t2 B m>s At y = 0, from Eq. [1] 0 = 27 – 3t2 t = 3. 00 s Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result y = 27 – 3t2 and applying Eq. 12–1, we have A+TB s 0 L ds = ydt t 0 L ds = A 27 – 3t2 B dt [2] s = A 27t – t3 B m At t = 3. 00 s, from Eq. [2] s = 27(3. 00) – 3. 003 = 54. 0 m Ans. •12–13.

A particle travels along a straight line such that in 2 s it moves from an initial position sA = +0. 5 m to a position sB = -1. 5 m. Then in another 4 s it moves from sB to sC = +2. 5 m. Determine the particle’s average velocity and average speed during the 6-s time interval. ?s = (sC – sA) = 2 m sT = (0. 5 + 1. 5 + 1. 5 + 2. 5) = 6 m t = (2 + 4) = 6 s vavg = ? s 2 = = 0. 333 m>s t 6 sT 6 = = 1 m>s t 6 Ans. (vsp)avg = Ans. 7 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 8 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–14. 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. Average Velocity: The displacement from A to C is ? s = sC – SA = -6 – (-8) = 2 m. yavg = 2 ? s = = 0. 222 m>s ? t 4 + 5 Ans. Average Speed: The distances traveled from A to B and B to C are sA : B = 8 + 3 = 11. m and sB : C = 3 + 6 = 9. 00 m, respectively. Then, the total distance traveled is sTot = sA : B + sB : C = 11. 0 – 9. 00 = 20. 0 m. A ysp B avg = sTot 20. 0 = = 2. 22 m>s ? t 4 + 5 Ans. 12–15. Tests reveal that a normal driver takes about 0. 75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0. 1% alcohol in his system to do the same. If such drivers are traveling on a straight road at 30 mph (44 ft>s) and their cars can decelerate at 2 ft>s2, determine the shortest stopping distance d for each from the moment they see the pedestrians.

Moral: If you must drink, please don’t drive! v1 44 ft/s d Stopping Distance: For normal driver, the car moves a distance of d? = yt = 44(0. 75) = 33. 0 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d? = 33. 0 ft and y = 0. + A:B y2 = y2 + 2ac (s – s0) 0 02 = 442 + 2(-2)(d – 33. 0) d = 517 ft Ans. For a drunk driver, the car moves a distance of d? = yt = 44(3) = 132 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d? = 132 ft and y = 0. + A:B y2 = y2 + ac (s – s0) 0 02 = 442 + 2(-2)(d – 132) d = 616 ft Ans. 8 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 9 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. *12–16. As a train accelerates uniformly it passes successive kilometer marks while traveling at velocities of 2 m>s and then 10 m>s. Determine the train’s velocity when it passes the next kilometer mark and the time it takes to travel the 2-km distance.

Kinematics: For the first kilometer of the journey, v0 = 2 m>s, v = 10 m>s, s0 = 0, and s = 1000 m. Thus, + A:B v2 = v0 2 + 2ac (s – s0) 102 = 22 + 2ac (1000 – 0) ac = 0. 048 m>s2 For the second 0. 048 m>s2. Thus, + A:B kilometer, v0 = 10 m>s, s0 = 1000 m, s = 2000 m, and v2 = v0 2 + 2ac (s – s0) v2 = 102 + 2(0. 048)(2000 – 1000) v = 14 m>s Ans. For the whole journey, v0 = 2 m>s, v = 14 m>s, and 0. 048 m>s2. Thus, + A:B v = v0 + act 14 = 2 + 0. 048t t = 250 s Ans. •12–17. A ball is thrown with an upward velocity of 5 m>s from the top of a 10-m high building.

One second later another ball is thrown vertically from the ground with a velocity of 10 m>s. Determine the height from the ground where the two balls pass each other. Kinematics: First, we will consider the motion of ball A with (vA)0 = 5 m>s, (sA)0 = 0, sA = (h – 10) m, tA = t? , and ac = -9. 81 m>s2. Thus, A+cB sA = (sA)0 + (vA)0 tA + h – 10 = 0 + 5t? + 1 actA 2 2 1 (-9. 81)(t? )2 2 (1) h = 5t? – 4. 905(t? )2 + 10 Motion of ball B is with (vB)0 = 10 m>s, (sB)0 = 0, sB = h, tB = t? – 1 and ac = -9. 81 m>s2. Thus, A+cB sB = (sB)0 + (vB)0 tB + h = 0 + 10(t? – 1) + ac tB 2 2 1 (-9. 81)(t? – 1)2 2 (2) h = 19. 81t? – 4. 905(t? )2 – 14. 905 Solving Eqs. (1) and (2) yields h = 4. 54 m t? = 1. 68 m 9 Ans. 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 10 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–18. A car starts from rest and moves with a constant acceleration of 1. 5 m>s2 until it achieves a velocity of 25 m>s.

It then travels with constant velocity for 60 seconds. Determine the average speed and the total distance traveled. Kinematics: For stage (1) of the motion, v0 = 0, s0 = 0, v = 25 m>s, and ac = 1. 5 m>s2. + A:B v = v0 + act 25 = 0 + 1. 5t1 t1 = 16. 67 s + A:B v2 = v0 2 + 2ac(s – s0) 252 = 0 + 2(1. 5)(s1 – 0) s1 = 208. 33 m For stage (2) of the motion, s0 = 108. 22 ft, v0 = 25 ft>s, t = 60 s, and ac = 0. Thus, + A:B s = s0 + v0t + 1 a t2 2 c s = 208. 33 + 25(60) + 0 = 1708. 33ft = 1708 m The average speed of the car is then vavg = s 1708. 33 = = 22. 3 m>s t1 + t2 16. 67 + 60 Ans. Ans. 12–19.

A car is to be hoisted by elevator to the fourth floor of a parking garage, which is 48 ft above the ground. If the elevator can accelerate at 0. 6 ft>s2, decelerate at 0. 3 ft>s2, and reach a maximum speed of 8 ft>s, determine the shortest time to make the lift, starting from rest and ending at rest. +c v2 = v2 + 2 ac (s – s0) 0 v2 = 0 + 2(0. 6)(y – 0) max 0 = v2 + 2(-0. 3)(48 – y) max 0 = 1. 2 y – 0. 6(48 – y) y = 16. 0 ft, +c vmax = 4. 382 ft>s 6 8 ft>s v = v0 + ac t 4. 382 = 0 + 0. 6 t1 t1 = 7. 303 s 0 = 4. 382 – 0. 3 t2 t2 = 14. 61 s t = t1 + t2 = 21. 9 s Ans. 10 91962_01_s12-p0001-0176 /8/09 8:06 AM Page 11 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. *12–20. A particle is moving along a straight line such that its speed 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. v = -4s2 ds = -4s2 dt s 2 L t 0 L s – 2 ds = -4 dt -s – 1| s = -4t|t 0 2 t = 1 -1 (s – 0. 5) 4 2 8t + 1 2 16 b = ab m>s 8t + 1 (8t + 1)2 s = v = -4 a a = Ans. 16(2)(8t + 1)(8) dv 256 = = a b m>s2 dt (8t + 1)4 (8t + 1)3 Ans. 11 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 12 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. •12–21. Two particles A and B start from rest at the origin s = 0 and move along a straight line such that aA = (6t – 3) ft>s2 and aB = (12t2 – 8) ft>s2, where t is in seconds.

Determine the distance between them when t = 4 s and the total distance each has traveled in t = 4 s. Velocity: The velocity of particles A and B can be determined using Eq. 12-2. dyA = aAdt yA t 0 L L 0 dyA = (6t – 3)dt yA = 3t2 – 3t dyB = aBdt yB 0 L t 0 L dyB = (12t2 – 8)dt yB = 4t3 – 8t The times when particle A stops are 4t3 – 8t = 0 t = 0 s and t = 22 s t = 0 s and = 1 s 3t2 – 3t = 0 The times when particle B stops are Position: The position of particles A and B can be determined using Eq. 12-1. dsA = yAdt sA 0 L t 0 L dsA = (3t2 – 3t)dt 3 2 t 2 sA = t3 dsB = yBdt sB t 0 L L 0 dsB = 4t3 – 8t)dt sB = t4 – 4t2 The positions of particle A at t = 1 s and 4 s are sA |t = 1 s = 13 sA |t = 4 s = 43 Particle A has traveled 3 2 (1 ) = -0. 500 ft 2 3 2 (4 ) = 40. 0 ft 2 The positions of particle B at t = 22 s and 4 s are dA = 2(0. 5) + 40. 0 = 41. 0 ft Ans. sB |t = 12 = (22)4 – 4(22)2 = -4 ft sB |t = 4 = (4)4 – 4(4)2 = 192 ft Particle B has traveled dB = 2(4) + 192 = 200 ft At t = 4 s the distance beween A and B is ? sAB = 192 – 40 = 152 ft 12 Ans. Ans. 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 13 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–22. A particle moving along a straight line is subjected to a deceleration a = (-2v3) m>s2, where v is in m>s. If it has a velocity v = 8 m>s and a position s = 10 m when t = 0, determine its velocity and position when t = 4 s. Velocity: The velocity of the particle can be related to its position by applying Eq. 12–3. ds = s 10m L ydy a dy – 2 8m>s 2y L 1 1 2y 16 8 16s – 159 [1] ds = s – 10 = y = Position: The position of the particle can be related to the time by applying Eq. 12–1. dt = t 0 L ds y s dt = 1 10m L 8 (16s – 159) ds 8t = 8s2 – 159s + 790 When t = 4 s, 8(4) = 8s2 – 159s + 790 8s2 – 159s + 758 = 0 Choose the root greater than 10 m s = 11. 94 m = 11. 9 m Substitute s = 11. 94 m into Eq. [1] yields y = 8 = 0. 250 m>s 16(11. 94) – 159 Ans. Ans. 13 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 14 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–23. 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. a = -2v dv = -2v dt dv -2 dt = L v 20 0 L ln v = -2t 20 v = (20e – 2t)m>s a = s 0 L v t Ans. Ans. dv = (-40e – 2t)m>s2 dt t 0 L t ds = v dt = 0 L (20e – 2t)dt s = -10e – 2t|t = -10(e – 2t – 1) 0 s = 10(1 – e – 2t)m Ans. 14 91962_01_s12-p0001-0176 /8/09 8:06 AM Page 15 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. *12–24. A particle starts from rest and travels along a straight line with an acceleration a = (30 – 0. 2v) ft>s2, where v is in ft>s. Determine the time when the velocity of the particle is v = 30 ft>s. Velocity: + A:B dt = t 0 L dv a dv 0 L 30 – 0. 2v v dt = t|t = 0 t = 5ln 1 ln(30 – 0. 2v) 2 0. 2 0 v 30 30 – 0. v t = 5ln 30 = 1. 12 s 30 – 0. 2(50) Ans. •12–25. When a particle is projected vertically upwards with an initial velocity of v0, it experiences an acceleration a = -(g + kv2) , where g is the acceleration due to gravity, k is a constant and v is the velocity of the particle. Determine the maximum height reached by the particle. Position: A+cB ds = s 0 L v dv a v v L0 ds = – s|s = – c 0 s = The particle achieves its maximum height when v = 0. Thus, hmax = g + kv0 2 1 ln ? ? g 2k g + kv0 2 1 ln ? ? 2k g + kv2 v 1 ln A g + kv2 B d 2 2k v0 vdv g + kv2 = 1 k ln ? 1 + v0 2 ? g 2k Ans. 15 91962_01_s12-p0001-0176 6/8/09 8:06 AM

Page 16 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–26. The acceleration of a particle traveling along a straight line is a = (0. 02et) m>s2, where t is in seconds. If v = 0, s = 0 when t = 0, determine the velocity and acceleration of the particle at s = 4 m. Velocity: a = 0. 02e5. 329 = 4. 13 m>s2 dv = a dt v t 0 L t A + : B Ans. Position: + A:B v|0 = 0. 02e 2 0 L v dv = 0. 02et dt v = C 0. 02 A et – 1 B D m>s 0 (1) ds = v dt s 0 L s t 0 L s = 0. 02 A et – t – 1 B m When s = 4 m, 4 = 0. 02 A et – t – 1 B et – t – 201 = 0 Solving the above equation by trial and error, t = 5. 329 s Thus, the velocity and acceleration when s = 4 m (t = 5. 329 s) are v = 0. 02 A e5. 329 – 1 B = 4. 11 m>s a = 0. 02e5. 329 = 4. 13 m>s2 Ans. Ans. s|0 = 0. 02 A e – t B 2 t ds = 0. 02 A et – 1 B dt t 0 12–27. A particle moves along a straight line with an acceleration of a = 5>(3s1>3 + s5>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 Simpson’s rule to evaluate the integral. 5 a = A 3s + s2 B 1 3 5 a ds = v dv 2 1 L A 3s + s 1 3 5 ds v 5 2 B = 0 L v dv 0. 8351 = 1 2 v 2 Ans. 16 v = 1. 29 m>s 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 17 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. *12–28. If the effects of atmospheric resistance are accounted for, a falling body has an acceleration defined by the equation a = 9. 1[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 ). Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2. (+ T) t 0 L dt = y dy a dt = y dy 2 0 L 9. 81[1 – (0. 01y) ] y t = dy dy 1 c + d 9. 81 L 2(1 + 0. 01y) 2(1 – 0. 01y) 0 0 L 9. 81t = 50lna 1 + 0. 01y b 1 – 0. 01y [1] y = 100(e0. 1962t – 1) e0. 1962t + 1 a) When t = 5 s, then, from Eq. [1] y = e0. 962t – 1 e0. 1962t + 1 100[e0. 1962(5) – 1] e0. 1962(5) + 1 = 45. 5 m>s Ans. b) If t : q , : 1. Then, from Eq. [1] ymax = 100 m>s Ans. 17 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 18 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. •12–29. The position of a particle along a straight line is given by s = (1. 5t3 – 13. 5t2 + 22. 5t) ft, where t is in seconds.

Determine the position of the particle when t = 6 s and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled. Position: The position of the particle when t = 6 s is s|t = 6s = 1. 5(63) – 13. 5(62) + 22. 5(6) = -27. 0 ft Ans. Total Distance Traveled: The velocity of the particle can be determined by applying Eq. 12–1. y = ds = 4. 50t2 – 27. 0t + 22. 5 dt The times when the particle stops are 4. 50t2 – 27. 0t + 22. 5 = 0 t = 1s and t = 5s The position of the particle at t = 0 s, 1 s and 5 s are s t = 0s = 1. 5(03) – 13. 5(02) + 22. 5(0) = 0 s t = 1s = 1. (13) – 13. 5(12) + 22. 5(1) = 10. 5 ft s t = 5s = 1. 5(53) – 13. 5(52) + 22. 5(5) = -37. 5 ft From the particle’s path, the total distance is stot = 10. 5 + 48. 0 + 10. 5 = 69. 0 ft Ans. 18 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 19 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–30. The velocity of a particle traveling along a straight line is v = v0 – ks, where k is constant.

If s = 0 when t = 0, determine the position and acceleration of the particle as a function of time. Position: + A:B dt = t 0 L ds y s 1 ln (v0 – ks) 2 k 0 dt = ds 0 L v0 – ks s tt = 0 t = ekt = s = Velocity: v = v0 1 ln ? ? k v0 – ks v0 v0 – ks v0 A 1 – e – kt B k Ans. d v0 ds = c A 1 – e – kt B d dt dt k v = v0e – kt Acceleration: a = d dv = A v e – kt B dt dt 0 Ans. a = -kv0e – kt 12–31. The acceleration of a particle as it moves along a straight line is given by a = 12t – 12 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. v t L 2 dv = 0 L (2 t – 1) dt v = t2 – t + 2 s 1 L t 0 L ds = s = (t2 – t + 2) dt 1 3 1 t – t2 + 2 t + 1 3 2 When t = 6 s, v = 32 m>s s = 67 m Since v Z 0 then d = 67 – 1 = 66 m Ans. Ans. Ans. 19 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 20 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–32. 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. Origin at roof: Ball A: A+cB s = s0 + v0 t + 1 a t2 2 c -s = 0 + 5t Ball B: 1 (9. 81)t2 2 A+cB s = s0 + v0 t + 1 a t2 2 c 1 (9. 81)t2 2 -s = -30 + 20t Solving, t = 2s s = 9. 62 m Distance from ground, d = (30 – 9. 62) = 20. 4 m Also, origin at ground, s = s0 + v0 t + 1 a t2 2 c 1 (-9. 81)t2 2 1 (-9. 81)t2 2 Ans. Ans. A = 30 + 5t + sB = 0 + 20t + Require sA = sB 30 + 5t + t = 2s sB = 20. 4 m 1 1 (-9. 81)t2 = 20t + (-9. 81)t2 2 2 Ans. Ans. 20 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 21 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. •12–33. A motorcycle starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft>s2 until it reaches a speed of 50 ft>s.

Afterwards it maintains this speed. Also, when t = 0, a car located 6000 ft down the road is traveling toward the motorcycle at a constant speed of 30 ft>s. Determine the time and the distance traveled by the motorcycle when they pass each other. Motorcycle: + A:B v = v0 + ac t? 50 = 0 + 6t? t? = 8. 33 s v2 = v2 + 2ac (s – s0) 0 (50)2 = 0 + 2(6)(s? – 0) s? = 208. 33 ft In t? = 8. 33 s car travels s– = v0 t? = 30(8. 33) = 250 ft Distance between motorcycle and car: 6000 – 250 – 208. 33 = 5541. 67 ft When passing occurs for motorcycle, s = v0 t; For car: s = v0 t; Solving, x = 3463. 4 ft t– = 69. 27 s Thus, for the motorcycle, t = 69. 27 + 8. 33 = 77. 6 s sm = 208. 33 + 3463. 54 = 3. 67(10)3 ft Ans. Ans. 5541. 67 – x = 30(t–) x = 50(t–) 21 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 22 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–34. A particle moves along a straight line with a velocity v = (200s) mm>s, where s is in millimeters.

Determine the acceleration of the particle at s = 2000 mm. How long does the particle take to reach this position if s = 500 mm when t = 0? Acceleration: + A:B Thus, a = v dv = 200s ds dv = (200s)(200) = 40 A 103 B s mm>s2 ds When s = 2000 mm, a = 40 A 103 B (2000) = 80 A 106 B mm>s2 = 80 km>s2 Position: + A:B dt = t Ans. ds v 1 lns 2 200 500 mm s t2 = 0 L t 0 dt = ds 500 L mm 200s s t = s 1 ln 200 500 At s = 2000 mm, t = 1 2000 ln = 6. 93 A 10 – 3 B s = 6. 93 ms 200 500 Ans. 22 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 23 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. ?12–35. A particle has an initial speed of 27 m>s. If it experiences a deceleration of a = 1-6t2 m>s2, where t is in seconds, determine its velocity, after it has traveled 10 m. How much time does this take? Velocity: + A:B dv = a dt t t v = (27 – 3t ) m>s + A:B ds = v dt s 0 L s t 0 L v2 27 L v dv = = 27 A -3t2 B 2 2 0 L (-6t)dt t 0 s = (27t – t ) m>s When s = 100 m, s 2 = A 27t – t3 B 2 0 3 ds =

A 27 – 3t2 B dt t 0 t = 0. 372 s v = 26. 6 m>s Ans. Ans. *12–36. The acceleration of a particle traveling along a straight line is a = (8 – 2s) m>s2, where s is in meters. If v = 0 at s = 0, determine the velocity of the particle at s = 2 m, and the position of the particle when the velocity is maximum. Velocity: + A:B v dv = a ds v 0 L s n = 216s – 2s2 m>s v s v2 ` = A 8s – s2 B 2 2 0 0 vdv = 0 L (8 – 2s) ds At s = 2 m, v s = 2 m = 216(2) – 2 A 22 B = ;4. 90 m>s dv 16 – 4s = = 0 ds 2 216s – 2s2 dv = 0. Thus, ds Ans. When the velocity is maximum 16 – 4s = 0 s = 4m 23 Ans. 91962_01_s12-p0001-0176 /8/09 8:07 AM Page 24 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. •12–37. Ball A is thrown vertically upwards with a velocity of v0. Ball B is thrown upwards from the same point with the same velocity t seconds later. Determine the elapsed time t 6 2v0>g from the instant ball A is thrown to when the balls pass each other, and find the velocity of each ball at this instant.

Kinematics: First, we will consider the motion of ball A with (vA)0 = v0, (sA)0 = 0, sA = h, tA = t? , and (ac)A = -g. A+cB sA = (sA)0 + (vA)0tA + h = 0 + v0t? + h = v0t? g 2 t? 2 1 (a ) t 2 2 cA A 1 ( -g)(t? )2 2 (1) A+cB vA = (vA)0 + (ac)A tA vA = v0 + (-g)(t? ) vA = v0 – gt? (2) The motion of ball B requires (vB)0 = v0, (sB)0 = 0, sB = h, tB = t? – t , and (ac)B = -g. A+cB sB = (sB)0 + (vB)0tB + h = 0 + v0(t? – t) + h = v0(t? – t) – 1 (a ) t 2 2 cBB 1 (-g)(t? – t)2 2 (3) g (t? – t)2 2 A+cB vB = (vB)0 + (ac)B tB vB = v0 + (-g)(t? – t) vB = v0 – g(t? – t) (4)

Solving Eqs. (1) and (3), g 2 g t? = v0(t? – t) – (t? – t)2 2 2 2v0 + gt t? = 2g v0t? Substituting this result into Eqs. (2) and (4), vA = v0 – ga = 2v0 + gt b 2g Ans. Ans. 1 1 gt = gt T 2 2 2v0 + gt – tb 2g vB = v0 – ga = 1 gt c 2 Ans. 24 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 25 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–38.

As a body is projected to a high altitude above the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formula a = -g0[R2>(R + y)2], where g0 is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If g0 = 9. 81 m>s2 and R = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth.

Hint: This requires that v = 0 as y : q . v dv = a dy 0 y L q v dv = -g0R 2 dy 2 g0 R2 q v2 2 0 2 = 2 y R + y 0 v = 22g0 R L (R + y) 0 = 22(9. 81)(6356)(10)3 Ans. = 11167 m>s = 11. 2 km>s 12–39. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12–38), derive an equation that relates the velocity of a freely falling particle to its altitude. Assume that the particle is released from rest at an altitude y0 from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude y0 = 500 km?

Use the numerical data in Prob. 12–38. From Prob. 12–38, (+ c ) a = -g0 R2 (R + y)2 Since a dy = v dv then y -g0 R2 g0 R2 c g0 R2[ Thus dy = 0 L v 2 y L0 (R + y) v dv y 1 v2 d = R + y y0 2 1 1 v2 ] = R + y R + y0 2 2g0 (y0 – y) A (R + y)(R + y0) v = -R When y0 = 500 km, 2(9. 81)(500)(103) v = -6356(103) A 6356(6356 + 500)(106) y = 0, Ans. v = -3016 m>s = 3. 02 km>s T 25 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 26 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. *12–40. 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. g dv = a = ? 2 ? A v2 – v2 B f dt vf v dy g t t = vf + v y g 1 ln ? ?` = 2t 2vf vf – v 0 vf vf 2g vf 2g ln ? ln ? vf + v vf – v vf + vf>2 vf – vf>2 b L v2 – v2? f = v2 f 0 L dt t = ? t = 0. 549 a vf g ? Ans. 26 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 27 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. •12–41. A particle is moving along a straight line such that its position from a fixed point is s = (12 – 15t2 + 5t3) m, where t is in seconds. Determine the total distance traveled by the particle from t = 1 s to t = 3 s.

Also, find the average speed of the particle during this time interval. Velocity: + A:B v = ds d = A 12 – 15t2 + 5t3 B dt dt v = -30t + 15t2 m>s The velocity of the particle changes direction at the instant when it is momentarily brought to rest. Thus, v = -30t + 15t2 = 0 t(-30 + 15t) = 0 t = 0 and 2 s Position: The positions of the particle at t = 0 s, 1 s, 2 s, and 3 s are s t = 0 s = 12 – 15 A 02 B + 5 A 03 B = 12 m s t = 2 s = 12 – 15 A 22 B + 5 A 23 B = -8 m s t = 3 s = 12 – 15 A 32 B + 5 A 33 B = 12 m s t = 1 s = 12 – 15 A 12 B + 5 A 13 B = 2 m Using the above results, the path of the particle is shown in Fig. . From this figure, the distance traveled by the particle during the time interval t = 1 s to t = 3 s is sTot = (2 + 8) + (8 + 12) = 30 m The average speed of the particle during the same time interval is vavg = sTot 30 = = 15 m>s ? t 3 – 1 Ans. Ans. 27 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 28 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–42.

The speed of a train during the first minute has been recorded as follows: t (s) v (m>s) 0 0 20 16 40 21 60 24 Plot the v-t graph, approximating the curve as straight-line segments between the given points. Determine the total distance traveled. The total distance traveled is equal to the area under the graph. sT = 1 1 1 (20)(16) + (40 – 20)(16 + 21) + (60 – 40)(21 + 24) = 980 m 2 2 2 Ans. 28 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 29 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–43. A two-stage missile is fired vertically from rest with the acceleration shown. In 15 s the first stage A burns out and the second stage B ignites. Plot the v-t and s-t graphs which describe the two-stage motion of the missile for 0 … t … 20 s. B a (m/s2) A 25 18 t (s) 15 20 a dt, the constant lines of the a–t graph become sloping lines for the v–t graph. L The numerical values for each point are calculated from the total area under the a–t graph to the point.

Since v = At t = 15 s, At t = 20 s, Since s = v = (18)(15) = 270 m>s v = 270 + (25)(20 – 15) = 395 m>s v dt, the sloping lines of the v–t graph become parabolic curves for the s–t graph. L The numerical values for each point are calculated from the total area under the v–t graph to the point. At t = 15 s, s = 1 (15)(270) = 2025 m 2 1 (395 – 270)(20 – 15) = 3687. 5 m = 3. 69 km 2 At t = 20 s, Also: 0 … t … 15: a = 18 s = 2025 + 270(20 – 15) + v = v0 + ac t = 0 + 18t s = s0 + v0 t + At t = 15: v = 18(15) = 270 s = 9(15)2 = 2025 15 … t … 20: a = 25 v = v0 + ac t = 270 + 25(t – 15) s = s0 + v0 t + When t = 20: v = 395 m>s s = 3687. m = 3. 69 km 29 1 1 a t2 = 2025 + 270(t – 15) + (25)(t – 15)2 2 c 2 1 a t2 = 0 + 0 + 9t2 2 c 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 30 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. *12–44. A freight train starts from rest and travels with a constant acceleration of 0. 5 ft>s2. After a time t? it maintains a constant speed so that when t = 160 s it has traveled 2000 ft.

Determine the time t? and draw the v–t graph for the motion. Total Distance Traveled: The distance for part one of the motion can be related to time t = t? by applying Eq. 12–5 with s0 = 0 and y0 = 0. + A:B s = s0 + y0 t + s1 = 0 + 0 + 1 a t2 2 c 1 (0. 5)(t? )2 = 0. 25(t? )2 2 The velocity at time t can be obtained by applying Eq. 12–4 with y0 = 0. + A:B y = y0 + act = 0 + 0. 5t = 0. 5t [1] The time for the second stage of motion is t2 = 160 – t? and the train is traveling at a constant velocity of y = 0. 5t? (Eq. [1]). Thus, the distance for this part of motion is + A:B s2 = yt2 = 0. t? (160 – t? ) = 80t? – 0. 5(t? )2 If the total distance traveled is sTot = 2000, then sTot = s1 + s2 2000 = 0. 25(t? )2 + 80t? – 0. 5(t? )2 0. 25(t? )2 – 80t? + 2000 = 0 Choose a root that is less than 160 s, then t? = 27. 34 s = 27. 3 s Ans. Y t Graph: The equation for the velocity is given by Eq. [1]. When t = t? = 27. 34 s, y = 0. 5(27. 34) = 13. 7 ft>s. 30 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 31 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. •12–45. If the position of a particle is defined by s = [2 sin (p>5)t + 4] m, where t is in seconds, construct the s-t, v-t, and a-t graphs for 0 … t … 10 s. 31 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 32 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–46.

A train starts from station A and for the first kilometer, it travels with a uniform acceleration. Then, for the next two kilometers, it travels with a uniform speed. Finally, the train decelerates uniformly for another kilometer before coming to rest at station B. If the time for the whole journey is six minutes, draw the v–t graph and determine the maximum speed of the train. For stage (1) motion, + A:B v1 = v0 + (ac)1 t vmax = 0 + (ac)1 t1 vmax = (ac)1t1 + A:B v1 2 = v0 2 + 2(ac)1(s1 – s0) vmax 2 = 0 + 2(ac)1(1000 – 0) (ac)1 = vmax 2 2000 (2) (1) Eliminating (ac)1 from Eqs. (1) and (2), we have t1 = 2000 vmax (3)

For stage (2) motion, the train travels with the constant velocity of vmax for t = (t2 – t1). Thus, + A:B s2 = s1 + v1t + 1 (a ) t2 2 c2 1000 + 2000 = 1000 + vmax (t2 – t1) + 0 t2 – t1 = 2000 vmax (4) For stage (3) motion, the train travels for t = 360 – t2. Thus, + A:B v3 = v2 + (ac)3t 0 = vmax – (ac)3(360 – t2) vmax = (ac)3(360 – t2) + A:B v3 2 = v2 2 + 2(ac)3(s3 – s2) 0 = vmax 2 + 2 C -(ac)3 D (4000 – 3000) (ac)3 = vmax 2 2000 (6) (5) Eliminating (ac)3 from Eqs. (5) and (6) yields 360 – t2 = Solving Eqs. (3), (4), and (7), we have t1 = 120 s t2 = 240 s Ans. 2000 vmax (7) vmax = 16. m>s Based on the above results the v -t graph is shown in Fig. a. 32 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 33 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. 12–47. The particle travels along a straight line with the velocity described by the graph. Construct the a-s graph. v (m/s) 13 10 v 4 s (m) 2s v 4 s 7 a s Graph: For 0 … s 6 3 m, a = v dv = (2s + 4)(2) = (4s + 8) m>s2 ds 6 + A:B At s = 0 m and 3 m, a|s = 0 m = 4(0) + 8 = 8 m>s2 a|s = 3 m = 4(3) + 8 = 20 m>s2 For 3m 6 s … 6 m, + A:B a = v dv = (s + 7)(1) = (s + 7) m>s2 ds At s = 3 m and 6 m, a|s = 3 m = 3 + 7 = 10 m>s2 a|s = 6 m = 6 + 7 = 13 m>s2 The a-s graph is shown in Fig. a. 33 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 34 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. *12–48.

The a–s graph for a jeep traveling along a straight road is given for the first 300 m of its motion. Construct the v–s graph. At s = 0, v = 0. a (m/s2) 2 200 300 s (m) a s Graph: The function of acceleration a in terms of s for the interval 0 m … s 6 200 m is a – 0 2 – 0 = s – 0 200 – 0 For the interval 200 m 6 s … 300 m, 0 – 2 a – 2 = s – 200 300 – 200 a = (-0. 02s + 6) m>s2 a = (0. 01s) m>s2 Y s Graph: The function of velocity y in terms of s can be obtained by applying ydy = ads. For the interval 0 m ? ss At s = 200 m, y = 0. 100(200) = 20. 0 m>s For the interval 200 m 6 s … 300 m, ydy = ads y 20. 0m>s L At s = 300 m, = 2-0. 02(3002) + 12(300) – 1200 = 24. 5 m>s y = A 2 -0. 02s2 + 12s – 1200 B m>s s ydy = 200m L (-0. 02s + 6)ds 34 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 35 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. 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. •12–49. A particle travels along a curve defined by the equation s = (t3 – 3t2 + 2t) m. where t is in seconds. Draw the s – t, v – t, and a – t graphs for the particle for 0 … t … 3 s. = t3 – 3t2 + 2t v = ds = 3t2 – 6t + 2 dt dv = 6t – 6 dt a = v = 0 at 0 = 3t2 – 6t + 2 t = 1. 577 s, and t = 0. 4226 s, s|t = 1. 577 = -0. 386 m s|t = 0. 4226 = 0. 385 m 12–50. A truck is traveling along the straight line with a velocity described by the graph. Construct the a-s graph for 0 … s … 1500 ft. a s Graph: For 0 … s 6 625 ft, a = v dv 3 = A 0. 6s3>4 B c (0. 6)s – 1>4 d = A 0. 27s1>2 B ft >s2 ds 4 v (ft/s) v 75 0. 6 s3/4 + A:B s(ft) 625 1500 At s = 625ft, a|s = 625 ft = 0. 27 A 6251>2 B = 6. 75ft>s2 For 625 ft 6 s 6 1500 ft, + A:B a = v dv = 75(0) = 0 ds The a-s graph is shown in Fig. a.