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# Chapter12 kinematic of a particle-dynamic 12th edition-solution

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### Chapter12 kinematic of a particle-dynamic 12th edition-solution

1. 1. 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 Ans. v = v0 + act 15 = 0 + 0.5625t t = 26.7 s 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 v = v0 + act Ans. = 0 + 1(30) = 30 m>s + A:B s = s0 + v0t + = 0 + 0 + 1 2 at 2 c 1 (1) A 302 B 2 = 450 m Ans. 1
2. 2. 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 A+TB Ans. 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 Ans. v = v0 + act 0 = 15 + (-2.25)t t = 6.67 s Ans. 2
3. 3. 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 t v dv = 0 L 0 L A 12t - 3t1>2 B dt vΗ0 = A 6t2 - 2t3>2 B 2 v 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, + A:B ds = v dt s t ds = 15 L ft 0 L A 6t2 - 2t3>2 B dt sΗ15 ft = a 2t3 s s = a 2t3 - 4 5>2 2 t t b 5 0 4 5>2 t + 15 bft 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 s = s0 + v0t + 1 a t2 2 c -h = 0 + 6(3) + 1 (-32.2) A 32 B 2 h = 127 ft A+cB Ans. v = v0 + act v = 6 + (-32.2)(3) = -90.6 ft>s = 90.6 ft>s Ans. T 3
4. 4. 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 Ans. v = 25 + (-3)(4) = 13 m>s ¢s = s - s0 = v0 t + 1 a t2 2 c ¢s = s - 0 = 25(4) + 1 (-3)(4)2 = 76 m 2 Ans. 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 s = s0 + v0 t + 1 a t2 2 c = 0 + 12(10) + 1 ( -2)(10)2 2 = 20 ft Ans. •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 = dv a t v dt = 0 L t 0 L dv k>v v L0 v dt = 1 vdv k v L0 t 1 2 2v t2 = v 2k v0 0 v = 22kt + v0 2 t = 1 A v2 - v0 2 B 2k Ans. 4
5. 5. 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 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 B s3 = yt1 = 60t1 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. 5
6. 6. 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 dv = -6t 2 = -24 m>s2 dt t=4 t ds = -10 L (1) 1 L t v dt = 1 L Ans. A 12 - 3t2 B dt s + 10 = 12t - t3 - 11 s = 12t - t3 - 21 s| t = 0 = -21 s|t = 10 = -901 Ans. ¢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. 6
7. 7. 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 dy = adt t y 27 L dy = 0 L -6tdt y = A 27 - 3t2 B m>s [1] 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 ds = ydt s 0 L t ds = 0 L 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. •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 (vsp)avg = Ans. sT 6 = = 1 m>s t 6 Ans. 7
8. 8. 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 = 11.0 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. v1 ϭ 44 ft/s 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! 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 + 2ac (s - s0) 0 02 = 442 + 2(-2)(d - 132) d = 616 ft Ans. 8
9. 9. 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 1 actA 2 2 sA = (sA)0 + (vA)0 tA + h - 10 = 0 + 5t¿ + 1 (-9.81)(t¿)2 2 h = 5t¿ - 4.905(t¿)2 + 10 (1) 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) + 1 ac tB 2 2 1 (-9.81)(t¿ - 1)2 2 h = 19.81t¿ - 4.905(t¿)2 - 14.905 (2) Solving Eqs. (1) and (2) yields h = 4.54 m Ans. t¿ = 1.68 m 9
10. 10. 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 Ans. The average speed of the car is then vavg = s 1708.33 = = 22.3 m>s t1 + t2 16.67 + 60 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
11. 11. 91962_01_s12-p0001-0176 6/8/09 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 s - 2 ds = 0 L -4 dt -s - 1| s = -4t|t 0 2 t = 1 -1 (s - 0.5) 4 s = 2 8t + 1 v = -4 a a = 2 2 16 b = ab m>s 8t + 1 (8t + 1)2 Ans. 16(2)(8t + 1)(8) dv 256 = = a b m>s2 dt (8t + 1)4 (8t + 1)3 Ans. 11
12. 12. 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 t yA dyA = L 0 (6t - 3)dt 0 L yA = 3t2 - 3t dyB = aBdt t yB dyB = 0 L 0 L (12t2 - 8)dt yB = 4t3 - 8t The times when particle A stops are 3t2 - 3t = 0 4t3 - 8t = 0 t = 0 s and t = 22 s t = 0 s and = 1 s 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 dsA = (3t2 - 3t)dt 0 L sA = t3 - 3 2 t 2 dsB = yBdt sB L 0 t dsB = 0 L (4t3 - 8t)dt sB = t4 - 4t2 The positions of particle A at t = 1 s and 4 s are sA |t = 1 s = 13 - 3 2 (1 ) = -0.500 ft 2 sA |t = 4 s = 43 - 3 2 (4 ) = 40.0 ft 2 Particle A has traveled 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 Ans. At t = 4 s the distance beween A and B is ¢sAB = 192 - 40 = 152 ft Ans. 12
13. 13. 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 ydy a y ds = 10m L s - 10 = y = dy - 2 8m>s 2y L 1 1 2y 16 8 16s - 159 [1] 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 Ans. Substitute s = 11.94 m into Eq. [1] yields y = 8 = 0.250 m>s 16(11.94) - 159 Ans. 13
14. 14. 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 t v dv -2 dt = L v 20 0 L ln v = -2t 20 v = (20e - 2t)m>s a = Ans. dv = (-40e - 2t)m>s2 dt Ans. s 0 L t ds = 0 L t v dt = 0 L (20e - 2t)dt s = -10e - 2t|t = -10(e - 2t - 1) 0 s = 10(1 - e - 2t)m Ans. 14
15. 15. 91962_01_s12-p0001-0176 6/8/09 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 dv a dt = t v dt = 0 L dv 0 L 30 - 0.2v 1 ln(30 - 0.2v) 2 0.2 0 v t|t = 0 t = 5ln t = 5ln 30 30 - 0.2v 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 = v dv a s 0 L v ds = s|s = - c 0 s = v L0 - vdv g + kv2 v 1 ln A g + kv2 B d 2 2k v0 g + kv0 2 1 ln ¢ ≤ 2k g + kv2 The particle achieves its maximum height when v = 0. Thus, hmax = = g + kv0 2 1 ln ¢ ≤ g 2k 1 k ln ¢ 1 + v0 2 ≤ g 2k Ans. 15
16. 16. 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 + : a = 0.02e5.329 = 4.13 m>s2 B Ans. dv = a dt t v dv = v|0 = 0.02e 2 0 L 0 L v 0.02et dt t t 0 v = C 0.02 A et - 1 B D m>s Position: + A:B (1) ds = v dt s t ds = 0.02 A et - 1 B dt s|0 = 0.02 A e - t B 2 0 L 0 L s t t 0 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 Ans. a = 0.02e5.329 = 4.13 m>s2 Ans. 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. a = 5 A 3s + s2 B 1 3 5 a ds = v dv 2 v 5 ds 1 L A 3s + s 1 3 0.8351 = 5 2 B = 0 L v dv 1 2 v 2 v = 1.29 m>s Ans. 16
17. 17. 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 = 9.81[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 ). Velocity: The velocity of the particle can be related to the time by applying Eq. 12–2. dt = (+ T) t 0 L dy a y dy 2 0 L 9.81[1 - (0.01y) ] dt = y t = y dy dy 1 c + d 9.81 L 2(1 + 0.01y) 2(1 - 0.01y) 0 0 L 9.81t = 50lna y = 1 + 0.01y b 1 - 0.01y 100(e0.1962t - 1) [1] e0.1962t + 1 a) When t = 5 s, then, from Eq. [1] y = b) If t : q , e0.1962t - 1 e0.1962t + 1 100[e0.1962(5) - 1] e0.1962(5) + 1 Ans. = 45.5 m>s : 1. Then, from Eq. [1] Ans. ymax = 100 m>s 17
18. 18. 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 sΗt = 0s = 1.5(03) - 13.5(02) + 22.5(0) = 0 sΗt = 1s = 1.5(13) - 13.5(12) + 22.5(1) = 10.5 ft sΗt = 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
19. 19. 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 ds y dt = t 0 L s dt = tΗt = 0 t = ds 0 L v0 - ks s 1 ln (v0 - ks) 2 k 0 v0 1 ln ¢ ≤ k v0 - ks ekt = v0 v0 - ks s = v0 A 1 - e - kt B k v = d v0 ds = c A 1 - e - kt B d dt dt k Ans. Velocity: v = v0e - kt Acceleration: a = d dv = A v e - kt B dt dt 0 a = -kv0e - kt Ans. 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. t v L 2 dv = (2 t - 1) dt 0 L v = t2 - t + 2 s 1 L t ds = s = 0 L (t2 - t + 2) dt 1 3 1 t - t2 + 2 t + 1 3 2 When t = 6 s, v = 32 m>s Ans. s = 67 m Ans. d = 67 - 1 = 66 m Ans. Since v Z 0 then 19
20. 20. 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 + -s = 0 + 5t - 1 a t2 2 c 1 (9.81)t2 2 Ball B: A+cB s = s0 + v0 t + 1 a t2 2 c -s = -30 + 20t - 1 (9.81)t2 2 Solving, t = 2s Ans. s = 9.62 m Distance from ground, d = (30 - 9.62) = 20.4 m Ans. Also, origin at ground, s = s0 + v0 t + 1 a t2 2 c sA = 30 + 5t + 1 (-9.81)t2 2 sB = 0 + 20t + 1 (-9.81)t2 2 Require sA = sB 30 + 5t + 1 1 (-9.81)t2 = 20t + (-9.81)t2 2 2 t = 2s Ans. sB = 20.4 m Ans. 20
21. 21. 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; x = 50(t–) For car: s = v0 t; 5541.67 - x = 30(t–) Solving, x = 3463.54 ft t– = 69.27 s Thus, for the motorcycle, t = 69.27 + 8.33 = 77.6 s Ans. sm = 208.33 + 3463.54 = 3.67(10)3 ft Ans. 21
22. 22. 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 dv = 200s ds Thus, a = v 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 Ans. Position: + A:B ds v dt = t s dt = t2 = 0 L t 0 t = ds 500 L mm 200s 1 lns 2 200 500 mm s 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
23. 23. 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. ½12–35. 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 27 L v2 t dv = v = 27 (-6t)dt 0 L A -3t2 B 2 t 0 2 v = (27 - 3t ) m>s + A:B ds = v dt s 0 L t ds = 0 L A 27 - 3t2 B dt s 2 = A 27t - t3 B 2 s 0 t 0 3 s = (27t - t ) m>s When s = 100 m, t = 0.372 s Ans. v = 26.6 m>s 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 vdv = 0 L (8 - 2s) ds v2 ` = A 8s - s2 B 2 2 0 0 n = 216s - 2s2 m>s s v At s = 2 m, vΗs = 2 m = 216(2) - 2 A 22 B = ;4.90 m>s Ans. dv = 0. Thus, ds dv 16 - 4s = = 0 ds 2 216s - 2s2 When the velocity is maximum 16 - 4s = 0 s = 4m Ans. 23
24. 24. 91962_01_s12-p0001-0176 6/8/09 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 h = 0 + v0t¿ + h = v0t¿ - A+cB 1 (a ) t 2 2 cA A sA = (sA)0 + (vA)0tA + 1 ( -g)(t¿)2 2 g 2 t¿ 2 (1) 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) - A+cB 1 (a ) t 2 2 cBB 1 (-g)(t¿ - t)2 2 g (t¿ - t)2 2 (3) 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¿ - Ans. Substituting this result into Eqs. (2) and (4), vA = v0 - ga = - 1 1 gt = gt T 2 2 vB = v0 - ga = 2v0 + gt b 2g Ans. 2v0 + gt - tb 2g 1 gt c 2 Ans. 24
25. 25. 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 q dy 2 y L v dv = -g0R L (R + y) 0 2 g0 R2 q v2 2 0 2 = 2 y R + y 0 v = 22g0 R = 22(9.81)(6356)(10)3 = 11167 m>s = 11.2 km>s Ans. 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, a = -g0 (+ c ) R2 (R + y)2 Since a dy = v dv then y -g0 R2 v dy 2 y L0 (R + y) = 0 L v dv g0 R2 c y 1 v2 d = R + y y0 2 g0 R2[ 1 1 v2 ] = R + y R + y0 2 Thus v = -R 2g0 (y0 - y) A (R + y)(R + y0) 2(9.81)(500)(103) v = -6356(103) A 6356(6356 + 500)(106) When y0 = 500 km, y = 0, v = -3016 m>s = 3.02 km>s T Ans. 25
26. 26. 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 t g dy 0 L v2 - v2¿ f = v2 f 0 L dt vf + v y g 1 ln ¢ ≤` = 2t 2vf vf - v 0 vf t = t = vf 2g vf 2g ln ¢ ln ¢ t = 0.549a vf + v vf - v ≤ vf + vf>2 vf - vf>2 vf g b ≤ Ans. 26
27. 27. 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 sΗt = 0 s = 12 - 15 A 02 B + 5 A 03 B = 12 m sΗt = 1 s = 12 - 15 A 12 B + 5 A 13 B = 2 m sΗt = 2 s = 12 - 15 A 22 B + 5 A 23 B = -8 m sΗt = 3 s = 12 - 15 A 32 B + 5 A 33 B = 12 m Using the above results, the path of the particle is shown in Fig. a. 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 Ans. The average speed of the particle during the same time interval is vavg = sTot 30 = = 15 m>s ¢t 3 - 1 Ans. 27
28. 28. 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) 0 20 40 60 v (m>s) 0 16 21 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 28 Ans.
29. 29. 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 a (m/s2) 25 18 t (s) 15 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, v = (18)(15) = 270 m>s At t = 20 s, v = 270 + (25)(20 - 15) = 395 m>s Since 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. 1 (15)(270) = 2025 m 2 At t = 15 s, s = At t = 20 s, s = 2025 + 270(20 - 15) + 1 (395 - 270)(20 - 15) = 3687.5 m = 3.69 km 2 Also: 0 … t … 15: a = 18 v = v0 + ac t = 0 + 18t s = s0 + v0 t + 1 a t2 = 0 + 0 + 9t2 2 c 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 + 1 1 a t2 = 2025 + 270(t - 15) + (25)(t - 15)2 2 c 2 When t = 20: v = 395 m>s s = 3687.5 m = 3.69 km 29 20
30. 30. 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 = 0.5t¿(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
31. 31. 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
32. 32. 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 (1) v1 2 = v0 2 + 2(ac)1(s1 - s0) vmax 2 = 0 + 2(ac)1(1000 - 0) (ac)1 = vmax 2 2000 (2) 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 1 (a ) t2 2 c2 s2 = s1 + v1t + 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 (5) 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) Eliminating (ac)3 from Eqs. (5) and (6) yields 360 - t2 = 2000 vmax (7) Solving Eqs. (3), (4), and (7), we have t1 = 120 s t2 = 240 s vmax = 16.7 m>s Ans. Based on the above results the v -t graph is shown in Fig. a. 32
33. 33. 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. v (m/s) 12–47. The particle travels along a straight line with the velocity described by the graph. Construct the a-s graph. 13 vϭsϩ7 10 v ϭ 2s ϩ 4 4 s (m) a؊s Graph: For 0 … s 6 3 m, + A:B a = v 3 dv = (2s + 4)(2) = (4s + 8) m>s2 ds 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 6
34. 34. 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 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 a = (0.01s) m>s2 For the interval 200 m 6 s … 300 m, 0 - 2 a - 2 = s - 200 300 - 200 a = (-0.02s + 6) 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 ◊ s<200 m, ydy = ds s y L 0 ydy = 0 L 0.01sds y = (0.1s) m>s At s = 200 m, y = 0.100(200) = 20.0 m>s For the interval 200 m 6 s … 300 m, ydy = ads y A 2 -0.02s2 + 12s - 1200 B m>s 20.0m>s L 200m L (-0.02s + 6)ds y = 2-0.02(3002) + 12(300) - 1200 = 24.5 m>s y = At s = 300 m, s ydy = 34 300 s (m)
35. 35. 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. s = t3 - 3t2 + 2t v = ds = 3t2 - 6t + 2 dt a = dv = 6t - 6 dt 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. v (ft/s) v ϭ 0.6 s3/4 a؊s Graph: For 0 … s 6 625 ft, + A:B a = v 75 dv 3 = A 0.6s3>4 B c (0.6)s - 1>4 d = A 0.27s1>2 B ft >s2 ds 4 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 s(ft) 625 a = v dv = 75(0) = 0 ds The a-s graph is shown in Fig. a. 35 1500
36. 36. 91962_01_s12-p0001-0176 6/8/09 8:09 AM Page 36 © 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. v(m/s) 12–51. A car starts from rest and travels along a straight road with a velocity described by the graph. Determine the total distance traveled until the car stops. Construct the s–t and a–t graphs. 30 v ϭ Ϫ0.5t ϩ 45 vϭt s؊t Graph: For the time interval 0 … t 6 30 s, the initial condition is s = 0 when t = 0 s. + A:B ds = vdt s 0 L t ds = tdt 0 L t2 s = ¢ ≤m 2 When t = 30 s, s = 302 = 450 m 2 For the time interval 30 s 6 t … 90 s, the initial condition is s = 450 m when t = 30 s. + A:B ds = vdt s t ds = (-0.5t + 45)dt 30s L 450 L m 1 s = a - t2 + 45t - 675 b m 4 When t = 90 s, sΗt = 90 s = - 1 A 902 B + 45(90) - 675 = 1350 m 4 Ans. The s - t graph shown is in Fig. a. a؊t Graph: For the time interval 0 6 t 6 30 s, a = dv d = (t) = 1 m>s2 dt dt For the time interval 30 s 6 t … 90 s, a = dv d = (-0.5t + 45) = -0.5 m>s2 dt dt The a- t graph is shown in Fig. b. Note: Since the change in position of the car is equal to the area under the v- t graph, the total distance traveled by the car is ¢s = L vdt sΗt = 90 s - 0 = 1 (90)(30) 2 sΗt = 90 s = 1350 s 36 t(s) 30 60 90
37. 37. 91962_01_s12-p0001-0176 6/8/09 8:09 AM Page 37 © 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. *12–52. A car travels up a hill with the speed shown. Determine the total distance the car travels until it stops (t = 60 s). Plot the a-t graph. v (m/s) 10 t (s) Distance traveled is area under v - t graph. s = (10)(30) + 30 1 (10)(30) = 450 m 2 60 Ans. v (m/s) •12–53. The snowmobile moves along a straight course according to the v–t graph. Construct the s–t and a–t graphs for the same 50-s time interval. When t = 0, s = 0. s؊t Graph: The position function in terms of time t can be obtained by applying ds 2 12 y = t = a t b m>s. . For time interval 0 s … t 6 30 s, y = dt 30 5 12 ds = ydt s 0 L t ds = 2 tdt 0 L 5 t (s) 30 1 s = a t2 b m 5 At t = 30 s , s = 1 A 302 B = 180 m 5 For time interval 30 s<t ◊ 50 s, ds = ydt s t ds = 180 L m 12dt 30 L s s = (12t - 180) m At t = 50 s, s = 12(50) - 180 = 420 m a؊t Graph: The acceleration function in terms of time t can be obtained by applying dy 2 dy a = = . For time interval 0 s ◊ t<30 s and 30 s<t ◊ 50 s, a = dt dt 5 dy = 0.4 m>s2 and a = = 0, respectively. dt 37 50
38. 38. 91962_01_s12-p0001-0176 6/8/09 8:09 AM Page 38 © 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. (vm)1 ϭ 60 ft/s 12–54. A motorcyclist at A is traveling at 60 ft>s when he wishes to pass the truck T which is traveling at a constant speed of 60 ft>s. To do so the motorcyclist accelerates at 6 ft>s2 until reaching a maximum speed of 85 ft>s. If he then maintains this speed, determine the time needed for him to reach a point located 100 ft in front of the truck. Draw the v-t and s-t graphs for the motorcycle during this time. (vm)2 ϭ 85 ft/s vt ϭ 60 ft/s A T 40 ft Motorcycle: Time to reach 85 ft>s, v = v0 + ac t 85 = 60 + 6t t = 4.167 s v2 = v2 + 2ac (s - s0) 0 Distance traveled, (85)2 = (60)2 + 2(6)(sm - 0) sm = 302.08 ft In t = 4.167 s, truck travels st = 60(4.167) = 250 ft Further distance for motorcycle to travel: 40 + 55 + 250 + 100 - 302.08 = 142.92 ft Motorcycle: s = s0 + v0 t (s + 142.92) = 0 + 85t¿ Truck: s = 0 + 60t¿ Thus t¿ = 5.717 s t = 4.167 + 5.717 = 9.88 s Ans. Total distance motorcycle travels sT = 302.08 + 85(5.717) = 788 ft 38 55 ft 100 ft
39. 39. 91962_01_s12-p0001-0176 6/8/09 8:09 AM Page 39 © 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. 12–55. An airplane traveling at 70 m>s lands on a straight runway and has a deceleration described by the graph. Determine the time t¿ and the distance traveled for it to reach a speed of 5 m>s. Construct the v–t and s-t graphs for this time interval, 0 … t … t¿ . a(m/s2) 5 Ϫ4 N؊t Graph: For the time interval 0 … t 6 5 s, the initial condition is v = 70 m>s when t = 0 s. + A:B dv = adt t v L m>s 70 dv = 0 L -10dt v = (-10t + 70) m>s When t = 5 s, v|t = 5 s = -10(5) + 70 = 20 m>s For the time interval 5 s 6 t … t¿ , the initial condition is v = 20 m>s when t = 5 s. + A:B dv = adt t v L m>s 20 dv = 5 Ls -4dt v = (-4t + 40) m>s When v = 5 m>s, t¿ = 8.75 s 5 = -4t¿ + 40 Ans. Also, the change in velocity is equal to the area under the a–t graph. Thus, ¢v = L adt 5 - 70 = - C 5(10) + 4(t¿ - 5) D t¿ = 8.75s¿ The v–t graph is shown in Fig. a. 39 Ϫ10 t¿ t(s)
40. 40. 91962_01_s12-p0001-0176 6/8/09 8:09 AM Page 40 © 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. 12–57. Continued s؊t Graph: For the time interval 0 … t 6 5 s, the initial condition is s = 0 when t = 0 s. + A:B ds = vdt s t ds = 0 L s = 0 L (-10t + 70)dt A -5t2 + 70t B m When t = 5 s, sΗt = 5 s = -5 A 52 B + 70(5) = 225 m For the time interval 5 6 t … t¿ = 8.75 s the initial condition is s = 225 m when t = 5 s. + A:B ds = vdt s t ds = 225 L m s = 5 L (-4t + 40)dt A -2t2 + 40t + 75 B m When t = t¿ = 8.75 s, sΗt = 8.75 s = -2 A 8.752 B + 40(8.75) + 75 = 271.875 m = 272 m Ans. Also, the change in position is equal to the area under the v–t graph. Referring to Fig. a, we have ¢s = L vdt sΗt = 8.75 s - 0 = 1 1 (70 + 20)(5) + (20 + 5)(3.75) = 271.875 m = 272 m 2 2 The s–t graph is shown in Fig. b. 40 Ans.
41. 41. 91962_01_s12-p0001-0176 6/8/09 8:10 AM Page 41 © 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. *12–56. The position of a cyclist traveling along a straight road is described by the graph. Construct the v–t and a–t graphs. s (m) 137.5 s ϭ Ϫ0.625 t2 ϩ 27.5t Ϫ 162.5 N؊t Graph: For the time interval 0 … t 6 10 s, + A:B v = ds d = A 0.05t3 B = A 0.15t2 B m>s dt dt 50 When t = 0 s and 10 s, vΗt = 0 = 0.15 A 0 2 B =0 vΗt = 10 s = 0.15 A 10 2 B = 15 m/s For the time interval 10 s 6 t … 20 s, + A:B v = d ds = A -0.625t2 + 27.5t - 162.5 B = ( -1.25t + 27.5) m>s dt dt When t = 10 s and 20 s, vΗt = 10 s = -1.25(10) + 27.5 = 15 m>s vΗt = 20 s = -1.25(20) + 27.5 = 2.5 m>s The v–t graph is shown in Fig. a. a؊t Graph: For the time interval 0 … t 6 10 s, + A:B a = dv d = A 0.15t2 B = (0.3t) m>s2 dt dt When t = 0 s and 10 s, aΗ t = 0 s = 0.3(0) = 0 aΗ t = 10 s = 0.3(10) = 3 m>s2 For the time interval 10 s 6 t … 20 s, + A:B a = dv d = (-1.25t + 27.5) = -1.25 m>s2 dt dt the a–t graph is shown in Fig. b. 41 s ϭ 0.05 t3 t (s) 10 20
42. 42. 91962_01_s12-p0001-0176 6/8/09 8:10 AM Page 42 © 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. •12–57. The dragster starts from rest and travels along a straight track with an acceleration-deceleration described by the graph. Construct the v-s graph for 0 … s … s¿, and determine the distance s¿ traveled before the dragster again comes to rest. a(m/s2) 25 a ϭ 0.1s ϩ 5 5 N ؊s Graph: For 0 … s 6 200 m, the initial condition is v = 0 at s = 0. + A:B 200 vdv = ads Ϫ15 v s vdv = 0 L 0 L (0.1s + 5)ds v2 2 = A 0.05s2 + 5s B 2 2 0 0 v = a 20.1s2 + 10sb m>s s v vΗs = 200 m = 20.1 A 2002 B + 10(200) = 77.46 m>s = 77.5 m>s At s = 200 m, For 200 m 6 s … s¿ , the initial condition is v = 77.46 m>s at s = 200 m. + A:B vdv = ads s v 77.46 L m>s v dv = 200 L m -15ds s v2 2 = -15sΗ200 m 2 77.46 m>s v v = A 2-30s + 12000 B m>s 0 = 2 -30s¿ + 12000 When v = 0, s¿ = 400 m Ans. The v–s graph is shown in Fig. a. 42 s¿ s (m)
43. 43. 91962_01_s12-p0001-0176 6/8/09 8:10 AM Page 43 © 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. 12–58. A sports car travels along a straight road with an acceleration-deceleration described by the graph. If the car starts from rest, determine the distance s¿ the car travels until it stops. Construct the v -s graph for 0 … s … s¿ . a(ft/s2) 6 1000 Ϫ4 N؊s Graph: For 0 … s 6 1000 ft, the initial condition is v = 0 at s = 0. + A:B vdv = ads v s vdv = 0 L 0 L 6ds A 212s1>2 B ft>s v2 = 6s 2 v = v = 212(1000)1>2 = 109.54 ft>s = 110 ft>s When s = 1000 ft, For 1000 ft 6 s … s¿ , the initial condition is v = 109.54 ft>s at s = 1000 ft. + A:B vdv = ads v s vdv = 109.54 ft>s L 1000 L ft -4ds s v2 2 = -4sΗ1000 ft 2 109.54 ft>s v v = A 220 000 - 8s B ft>s 0 = 220 000 - 8s¿ When v = 0, s¿ = 2500 ft Ans. The v–s graph is shown in Fig. a. 43 s¿ s(ft)
44. 44. 91962_01_s12-p0001-0176 6/8/09 8:10 AM Page 44 © 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. 12–59. A missile starting from rest travels along a straight track and for 10 s has an acceleration as shown. Draw the v -t graph that describes the motion and find the distance traveled in 10 s. a (m/s2) 40 a ϭ 2t ϩ 20 30 For t … 5 s, a ϭ 6t a = 6t dv = a dt t v dv = 0 L 0 L 6t dt t(s) 5 v = 3t2 When t = 5 s, v = 75 m>s For 5 6 t 6 10 s, a = 2t + 20 dv = a dt t v 75 L dv = 5 L (2t + 20) dt v - 75 = t2 + 20t - 125 v = t2 + 20t - 50 When t = 10 s, v = 250 m>s Distance at t = 5 s: ds = v dt s 0 L 5 ds = 0 L 3t2 dt s = (5)3 = 125 m Distance at t = 10 s: ds = v dv s 10 ds = 125 L s - 125 = 5 L A t2 + 20t - 50 B dt 10 1 3 t + 10t2 - 50t d 3 5 s = 917 m Ans. 44 10
45. 45. 91962_01_s12-p0001-0176 6/8/09 8:10 AM Page 45 © 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. *12–60. A motorcyclist starting from rest travels along a straight road and for 10 s has an acceleration as shown. Draw the v-t graph that describes the motion and find the distance traveled in 10 s. a (m/s2) dv = a dt For 0 … t 6 6 6 t v 1 2 dv = t dt 0 0 L L 6 v = t (s) 1 3 t 18 6 ds = v dt s t ds = 0 L s = When t = 6 s, 1 3 t dt 18 0 L 1 4 t 72 v = 12 m>s s = 18 m dv = a dt For 6 6 t … 10 t v 12 L dv = 6 dt 6 L v = 6t - 24 ds = v dt s 18 L t ds = 6 L (6t - 24) dt s = 3t2 - 24t + 54 When t = 10 s, 1 a ϭ — t2 6 v = 36 m>s s = 114 m Ans. 45 10
46. 46. 91962_01_s12-p0001-0176 6/8/09 8:19 AM Page 46 © 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. v (m/s) •12–61. The v -t graph of a car while traveling along a road is shown. Draw the s-t and a -t graphs for the motion. 20 0 … t … 5 20 ¢v = = 4 m>s2 ¢t 5 a = 5 … t … 20 a = 20 … t … 30 a = 5 20 20 - 20 ¢v = = 0 m>s2 ¢t 20 - 5 ¢v 0 - 20 = = -2 m>s2 ¢t 30 - 20 From the v–t graph at t1 = 5 s, t2 = 20 s, and t3 = 30 s, s1 = A1 = 1 (5)(20) = 50 m 2 s2 = A1 + A2 = 50 + 20(20 - 5) = 350 m s3 = A1 + A2 + A3 = 350 + 1 (30 - 20)(20) = 450 m 2 The equations defining the portions of the s–t graph are s 0 … t … 5s v = 4t; ds = v dt; 0 L t ds = 0 L s 5 … t … 20 s v = 20; ds = v dt; 50 L ds = s = 2t2 4t dt; t 5 L 20 dt; s = 20t - 50 s 20 … t … 30 s v = 2(30 - t); ds = v dt; 350 L t ds = 20 L 46 2(30 - t) dt; s = -t2 + 60t - 450 30 t (s)
47. 47. 91962_01_s12-p0001-0176 6/8/09 8:20 AM Page 47 © 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. 12–62. The boat travels in a straight line with the acceleration described by the a -s graph. If it starts from rest, construct the v-s graph and determine the boat’s maximum speed. What distance s¿ does it travel before it stops? a(m/s2) 6 a ϭ Ϫ0.02s ϩ 6 3 N؊s Graph: For 0 … s 6 150 m, the initial condition is v = 0 at s = 0. + A:B vdv = ads 150 s v vdv = 0 L v2 2 = 2 0 0 L ( -0.02s + 6)ds Ϫ4 v = a 2 -0.02s2 + 12sb m>s v A -0.01s2 + 6s B Η0 s vmax = vΗs = 150 m = 2 -0.02 A 1502 B + 12(150) = 36.74 m>s = 36.7 m>s The maximum velocity of the boat occurs at s = 150 m, where its acceleration changes sign. Thus, Ans. For 150 m 6 s 6 s¿ , the initial condition is v = 36.74 m>s at s = 150 m. + A:B vdv = ads v s vdv = 36.74 L m>s 150 L m -4ds v2 2 = -4s 2 2 36.74 m>s 150 m v = 2 -8s + 2550 m>s v s 0 = 2 -8s¿ + 2550 Thus, when v = 0, s¿ = 318.7 m = 319 m The v–s graph is shown in Fig. a. 47 Ans. s¿ s(m)
48. 48. 91962_01_s12-p0001-0176 6/8/09 8:20 AM Page 48 © 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. 12–63. The rocket has an acceleration described by the graph. If it starts from rest, construct the v-t and s-t graphs for the motion for the time interval 0 … t … 14 s. a(m/s2) N؊t Graph: For the time interval 0 … t 6 9 s, the initial condition is v = 0 at s = 0. A+cB 38 dv = adt a2 ϭ 36t a ϭ 4t Ϫ 18 t v dv = 0 L 0 L 1>2 6t dt 18 v = A 4t3>2 B m>s t(s) When t = 9 s, 9 vΗt = 9 s = 4 A 93>2 B = 108 m>s The initial condition is v = 108 m>s at t = 9 s. A+cB dv = adt t v dv = 9 Ls 108 L m>s (4t - 18)dt v = A 2t2 - 18t + 108 B m>s When t = 14 s, vΗt = 14 s = 2 A 142 B - 18(14) + 108 = 248 m>s The v–t graph is shown in Fig. a. s؊t Graph: For the time interval 0 … t 6 9 s, the initial condition is s = 0 when t = 0. A+cB ds = vdt s t ds = 0 L s = 0 L 4t3>2 dt 8 5>2 t 5 When t = 9 s, sΗt = 9 s = 8 5>2 A 9 B = 388.8 m 5 For the time interval 9 s 6 t … 14 s, the initial condition is s = 388.8 m when t = 9 s. A+cB ds = vdt A 2t2 - 18t + 108 B dt 388.8 9 L m Ls 2 s = a t3 - 9t2 + 108t - 340.2b m 3 s t ds = When t = 14 s, sΗt = 14 s = 2 A 143 B - 9 A 142 B + 108(14) - 340.2 = 1237 m 3 The s–t graph is shown in Fig. b. 48 14
49. 49. 91962_01_s12-p0001-0176 6/8/09 8:20 AM Page 49 © 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. v(m/s) *12–64. The jet bike is moving along a straight road with the speed described by the v -s graph. Construct the a-s graph. v ϭ 5s1/2 75 v ϭ Ϫ0.2s ϩ 120 15 a ؊ s Graph: For 0 … s 6 225 m, + A:B a = v s (m) 225 dv 5 = A 5s1>2 B a s - 1>2 b = 12.5 m>s2 ds 2 For 225 m 6 s … 525 m, + A:B a = v dv = (-0.2s + 120)(-0.2) = (0.04s - 24) m>s2 ds At s = 225 m and 525 m, aΗs = 225 m = 0.04(225) - 24 = -15 m>s2 aΗs = 525 m = 0.04(525) - 24 = -3 m>s2 The a–s graph is shown in Fig. a. 49 525
50. 50. 91962_01_s12-p0001-0176 6/8/09 8:20 AM Page 50 © 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. •12–65. The acceleration of the speed boat starting from rest is described by the graph. Construct the v-s graph. a(ft/s2) 10 a ϭ 0.04s ϩ 2 2 N ؊s Graph: For 0 … s 6 200 ft, the initial condition is v = 0 at s = 0. + A:B vdv = ads s v vdv = 0 L 0 L (0.04s + 2)ds s v2 2 = 0.02s2 + 2sΗ0 2 0 v = 20.04s2 + 4s ft>s v vΗs = 200 ft = 20.04 A 2002 B + 4(200) = 48.99 ft>s = 49.0 ft >s At s = 200 ft, For 200 ft 6 s … 500 ft, the initial condition is v = 48.99 ft>s at s = 200 ft. + A:B vdv = ads s v vdv = 48.99 L ft>s 10ds 200 L ft s v2 2 = 10sΗ200 ft 2 48.99 ft>s v = 220s - 1600 ft>s v vΗs = 500 ft = 220(500) - 1600 = 91.65 ft>s = 91.7 ft>s At s = 500 ft, The v–s graph is shown in Fig. a. 50 s(ft) 200 500
51. 51. 91962_01_s12-p0001-0176 6/8/09 8:21 AM Page 51 © 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. v(m/s) 12–66. The boat travels along a straight line with the speed described by the graph. Construct the s-t and a-s graphs. Also, determine the time required for the boat to travel a distance s = 400 m if s = 0 when t = 0. 80 s؊t Graph: For 0 … s 6 100 m, the initial condition is s = 0 when t = 0 s. A + : B v ϭ 0.2s ds dt = v t s v2 ϭ 4s ds dt = 0 0 L L 2s1>2 1>2 t = s 20 s(m) s = A t2 B m 100 When s = 100 m, 100 = t2 t = 10 s For 100 6 s … 400 m, the initial condition is s = 100 m when t = 10 s. + A:B ds v dt = t s dt = ds L s 10 100 L m 0.2s s t - 10 = 5ln 100 t s - 2 = ln 5 100 s et>5 - 2 = 100 et>5 s = 100 e2 s = A 13.53et>5 B m When s = 400 m, 400 = 13.53et>5 t = 16.93 s = 16.9 s Ans. The s–t graph is shown in Fig. a. a؊s Graph: For 0 m … s 6 100 m, a = v dv = A 2s1>2 B A s - 1>2 B = 2 m>s2 ds For 100 m 6 s … 400 m, a = v dv = (0.2s)(0.2) = 0.04s ds When s = 100 m and 400 m, aΗs = 100 m = 0.04(100) = 4 m>s2 aΗs = 400 m = 0.04(400) = 16 m>s2 The a–s graph is shown in Fig. b. 51 400
52. 52. 91962_01_s12-p0001-0176 6/8/09 8:21 AM Page 52 © 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. 12–67. The s–t graph for a train has been determined experimentally. From the data, construct the v-t and a–t graphs for the motion. s (m) 600 s ϭ 24t Ϫ 360 360 s ϭ 0.4t2 t (s) ds . Y؊t Graph: The velocity in terms of time t can be obtained by applying y = dt For time interval 0 s … t … 30 s, y = When t = 30 s, ds = 0.8t dt y = 0.8(30) = 24.0 m>s For time interval 30 s 6 t … 40 s, y = ds = 24.0 m>s dt a؊t Graph: The acceleration in terms of time t can be obtained by applying a = dy . dt dy = 0.800 m>s2 and For time interval 0 s … t 6 30 s and 30 s 6 t … 40 s, a = dt dy = 0, respectively. a = dt 52 30 40
53. 53. 91962_01_s12-p0001-0176 6/8/09 8:21 AM Page 53 © 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. *12–68. The airplane lands at 250 ft>s on a straight runway and has a deceleration described by the graph. Determine the distance s¿ traveled before its speed is decreased to 25 ft>s. Draw the s-t graph. a(ft/s2) 1750 s¿ s(ft) N؊s Graph: For 0 … s 6 1750 ft, the initial condition is v = 250 ft>s at s = 0 s. Ϫ7.5 + A:B Ϫ15 vdv = ads v s vdv = -15ds 250 0 L ft>s L s v2 2 = -15sΗ0 2 250 ft>s v v = A 262 500 - 30s B ft>s vΗs = 1750 ft = 262 500 - 30(1750) = 100 ft>s At s = 1750 ft, For 1750 ft 6 s 6 s¿ , the initial condition is v = 100 ft>s at s = 1750 ft. + A:B vdv = ads s v vdv = 100 L ft>s 1750 L ft -7.5ds s v2 2 = (-7.5s)Η1750 ft 2 100 ft>s v = 236 250 - 15s v 25 = 236 250 - 15s When v = 25 ft>s s¿ = 2375 ft Ans. The v–s graph is shown in Fig. a. 53
54. 54. 91962_01_s12-p0001-0176 6/8/09 8:22 AM Page 54 © 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. •12–69. The airplane travels along a straight runway with an acceleration described by the graph. If it starts from rest and requires a velocity of 90 m>s to take off, determine the minimum length of runway required and the time t¿ for take off. Construct the v -t and s-t graphs. a(m/s2) 8 v؊t graph: For the time interval 0 … t 6 10 s, the initial condition is v = 0 when t = 0 s. A + : B t(s) 10 dv = adt t v L 0 dv = 0 L 0.8tdt v = A 0.4t2 B m>s When t = 10 s, v = 0.4 A 102 B = 40 m>s For the time interval 10 s 6 t … t¿ , the initial condition is v = 40 m>s when t = 10 s. + A:B a ϭ 0.8t dv = adt t v dv = 8dt 10 L s 40 L m>s t v vΗ40 m>s = 8tΗ10 s v = (8t - 40) m>s Thus, when v = 90 m>s, 90 = 8t¿ - 40 t¿ = 16.25 s Ans. Also, the change in velocity is equal to the area under the a-t graph. Thus, ¢v = L adt 1 (8)(10) + 8(t¿ - 10) 2 t¿ = 16.25 s 90 - 0 = Ans. The v–t graph is shown in Fig. a. 54 t¿
55. 55. 91962_01_s12-p0001-0176 6/8/09 8:22 AM Page 55 © 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. s؊t Graph: For the time interval 0 … t 6 10 s, the initial condition is s = 0 when t = 0 s. + A:B ds = vdt s L 0 t ds = 2 0 L 0.4t dt s = A 0.1333t3 B m When t = 10 s, sΗt = 10 s = 0.1333 A 103 B = 133.33m For the time interval 10 s 6 t … t¿ = 16.25 s, the initial condition is s = 133.33m when t = 10 s. + A:B ds = vdt s t ds = (8t - 40)dt 10s L sΗ133.33 m = A 4t2 - 40t B 2 133.33 m L s t 10 s s = A 4t - 40t + 133.33 B m 2 When t = t¿ = 16.25 s sΗt = 16.25 s = 4(16.25)2 - 40(16.25) + 133.33 = 539.58 m = 540 m Ans. The s–t graph is shown in Fig. b. 55
56. 56. 91962_01_s12-p0001-0176 6/8/09 8:22 AM Page 56 © 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. 12–70. The a–t graph of the bullet train is shown. If the train starts from rest, determine the elapsed time t¿ before it again comes to rest. What is the total distance traveled during this time interval? Construct the v–t and s–t graphs. a(m/s2) N؊t Graph: For the time interval 0 … t 6 30 s, the initial condition is v = 0 when t = 0 s. + A:B t v L 0 0 L 0.1tdt v = A 0.05t2 B m>s When t = 30 s, vΗt = 30 s = 0.05 A 302 B = 45 m>s For the time interval 30 s 6 t … t¿ , the initial condition is v = 45 m>s at t = 30 s. + A:B dv = adt t v dv = 45 L m>s v = ¢- Thus, when v = 0, 0 = - 30 L s ¢- 1 t + 5 ≤ dt 15 1 2 t + 5t - 75 ≤ m>s 30 1 2 t¿ + 5t¿ - 75 30 Choosing the root t¿ 7 75 s, t¿ = 133.09 s = 133 s Ans. Also, the change in velocity is equal to the area under the a–t graph. Thus, ¢v = 0 = L adt 1 1 1 (3)(75) + B ¢ - t¿ + 5 ≤ (t¿ - 75) R 2 2 15 0 = - 1 2 t¿ + 5t¿ - 75 30 56 1 a ϭ Ϫ( 15 )t ϩ 5 t¿ 30 dv = adt dv = a ϭ 0.1t 3 75 t(s)
57. 57. 91962_01_s12-p0001-0176 6/8/09 8:22 AM Page 57 © 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. This equation is the same as the one obtained previously. The slope of the v–t graph is zero when t = 75 s, which is the instant a = vΗt = 75 s = - dv = 0.Thus, dt 1 A 752 B + 5(75) - 75 = 112.5 m>s 30 The v–t graph is shown in Fig. a. s؊t Graph: Using the result of v, the equation of the s–t graph can be obtained by integrating the kinematic equation ds = vdt. For the time interval 0 … t 6 30 s, the initial condition s = 0 at t = 0 s will be used as the integration limit. Thus, + A:B ds = vdt s 0 L t ds = s = a 0 L 0.05t2 dt 1 3 t bm 60 When t = 30 s, 1 A 303 B = 450 m 60 sΗt = 30 s = For the time interval 30 s 6 t … t¿ = 133.09 s, the initial condition is s = 450 m when t = 30 s. + A:B ds = vdt s t ds = 450 L m 30 L s a- 1 2 t + 5t - 75bdt 30 5 1 s = a - t3 + t2 - 75t + 750b m 90 2 When t = 75 s and t¿ = 133.09 s, sΗt = 75 s = - 1 5 A 753 B + A 752 B - 75(75) + 750 = 4500 m 90 2 sΗt = 133.09 s = - 1 5 A 133.093 B + A 133.092 B - 75(133.09) + 750 = 8857 m 90 2 The s–t graph is shown in Fig. b. 57 Ans.
58. 58. 91962_01_s12-p0001-0176 6/8/09 8:23 AM Page 58 © 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. 12–71. The position of a particle is r = 5(3t 3 - 2t)i – (4t1/2 + t)j + (3t2 - 2)k6 m, where t is in seconds. Determine the magnitude of the particle’s velocity and acceleration when t = 2 s. Velocity: v = dr d = c A 3t3 - 2t B i - A 4t1>2 + t B j + A 3t2 - 2 B k d = c A 9t2 - 2 B i - A 2t - 1>2 + 1 B j + (6t)k d m>s dt dt When t = 2 s, v = B c9 A 22 B - 2 d i - c2 A 2 - 1>2 B + 1 dj + 6(2)k R m>s = [34i - 2.414j + 12k] m>s v = 2vx 2 + vy 2 + vz 2 = 2342 + (-2.414)2 + 122 = 36.1 m>s Thus, the magnitude of the particle’s velocity is Ans. Acceleration: a = dv d = c A 9t2 - 2 B i - A 2t - 1>2 + 1 B j + (6t)k d m>s = C (18t)i + t - 3>2j + 6k D m>s2 dt dt When t = 2 s, a = C 18(2)i + 2 - 3>2j + 6k D m>s2 = [36i + 0.3536j + 6k] m>s2 a = 2ax 2 + ay 2 + az 2 = 2362 + 0.35362 + 62 = 36.5 m>s2 Thus, the magnitude of the particle’s acceleration is Ans. *12–72. The velocity of a particle is v = 53i + (6 - 2t)j6 m>s, where t is in seconds. If r = 0 when t = 0, determine the displacement of the particle during the time interval t = 1 s to t = 3 s. Position: The position r of the particle can be determined by integrating the kinematic equation dr = vdt using the initial condition r = 0 at t = 0 as the integration limit. Thus, dr = vdt t r 0 L dr = 0 L C 3i + (6 - 2t)j D dt r = c 3ti + A 6t - t2 B j dm When t = 1 s and 3 s, rΗt = 1 s = 3(1)i + C 6(1) - 12 D j = [3i + 5j] m>s rΗt = 3 s = 3(3)i + C 6(3) - 32 D j = [9i + 9j] m>s Thus, the displacement of the particle is ¢r = rΗt = 3 s - rΗt = 1 s = (9i + 9j) - (3i + 5j) = [6i + 4j] m Ans. 58
59. 59. 91962_01_s12-p0001-0176 6/8/09 8:23 AM Page 59 © 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. •12–73. A particle travels along the parabolic path y = bx2. If its component of velocity along the y axis is vy = ct2, determine the x and y components of the particle’s acceleration. Here b and c are constants. Velocity: dy = vy dt t y 0 L dy = y = 0 L ct2 dt c 3 t 3 Substituting the result of y into y = bx2, c 3 t = bx2 3 x = c 3>2 t A 3b Thus, the x component of the particle’s velocity can be determined by taking the time derivative of x. d c 3>2 3 c 1>2 # vx = x = t R = t B dt A 3b 2A 3b d 3 c 1>2 3 c - 1>2 3 c 1 # t ≤ = t = ax = vx = ¢ dt 2A 3b 4A 3b 4A 3b 2t Acceleration: d 2 # ay = vy = A ct B = 2ct dt 59 Ans. Ans.
60. 60. 91962_01_s12-p0001-0176 6/8/09 8:24 AM Page 60 © 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. 12–74. The velocity of a particle is given by v = 516t2i + 4t3j + (5t + 2)k6 m>s, where t is in seconds. If the particle is at the origin when t = 0, determine the magnitude of the particle’s acceleration when t = 2 s. Also, what is the x, y, z coordinate position of the particle at this instant? Acceleration: The acceleration expressed in Cartesian vector form can be obtained by applying Eq. 12–9. a = dv = {32ti + 12t2j + 5k} m>s2 dt When t = 2 s, a = 32(2)i + 12 A 22 B j + 5k = {64i + 48j + 5k} m>s2. The magnitude of the acceleration is a = 2a2 + a2 + a2 = 2642 + 482 + 52 = 80.2 m>s2 x y z Ans. Position: The position expressed in Cartesian vector form can be obtained by applying Eq. 12–7. dr = v dt t r 0 L dr = r = c 0 L A 16t2i + 4t3j + (5t + 2)k B dt 16 3 5 t i + t4j + a t2 + 2t bk d m 3 2 When t = 2 s, r = 16 3 5 A 2 B i + A 24 B j + c A 22 B + 2(2) dk = {42.7i + 16.0j + 14.0k} m. 3 2 Thus, the coordinate of the particle is (42.7, 16.0, 14.0) m Ans. 60
61. 61. 91962_01_s12-p0001-0176 6/8/09 8:24 AM Page 61 © 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. 12–75. A particle travels along the circular path x2 + y2 = r2. If the y component of the particle’s velocity is vy = 2r cos 2t, determine the x and y components of its acceleration at any instant. Velocity: dy = vy dt t y 0 L dy = 0 L 2r cos 2tdt y = r sin 2t Substituting this result into x2 + y2 = r2, we obtain x2 + r2 sin2 2t = r2 x2 = r2 A 1 - sin2 2t B x = <r cos 2t Thus, d # (<r cos 2t) = <2r sin 2t vx = x = dt Acceleration: d # (<2r sin 2t) = <4r cos 2t ax = vx = dt Ans. d # ay = vy = (2r cos 2t) = -4r sin 2t dt Ans. 61
62. 62. 91962_01_s12-p0001-0176 6/8/09 8:24 AM Page 62 © 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. *12–76. The box slides down the slope described by the equation y = (0.05x2) m, where x is in meters. If the box has x components of velocity and acceleration of vx = –3 m>s and ax = –1.5 m>s2 at x = 5 m, determine the y components of the velocity and the acceleration of the box at this instant. y y ϭ 0.05 x2 x Velocity: The x and y components of the box’s velocity can be related by taking the first time derivative of the path’s equation using the chain rule. y = 0.05x2 # # y = 0.1xx or vy = 0.1xvx At x = 5 m, vx = -3 m>s. Thus, vy = 0.1(5)(-3) = -1.5 m>s = 1.5 m>s T Ans. Acceleration: The x and y components of the box’s acceleration can be obtained by taking the second time derivative of the path’s equation using the chain rule. # # # y = 0.1[xx + xx] = 0.1 A x2 + xx B or ay = 0.1 A vx 2 + xax B At x = 5 m, vx = -3 m>s and ax = -1.5 m>s2. Thus, ay = 0.1 C (-3)2 + 5(-1.5) D = 0.15 m>s2 c Ans. 62
63. 63. 91962_01_s12-p0001-0176 6/8/09 8:24 AM Page 63 © 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. •12–77. The position of a particle is defined by r = 55 cos 2t i + 4 sin 2t j6 m, where t is in seconds and the arguments for the sine and cosine are given in radians. Determine the magnitudes of the velocity and acceleration of the particle when t = 1 s. Also, prove that the path of the particle is elliptical. Velocity: The velocity expressed in Cartesian vector form can be obtained by applying Eq. 12–7. v = dr = {-10 sin 2ti + 8 cos 2tj} m>s dt y = 2y2 + y2 = 2( -9.093)2 + (-3.329)2 = 9.68 m>s x y When t = 1 s, v = -10 sin 2(1)i + 8 cos 2(1)j = {-9.093i - 3.329j} m>s. Thus, the magnitude of the velocity is Ans. Acceleration: The acceleration expressed in Cartesian vector from can be obtained by applying Eq. 12–9. a = dv = {-20 cos 2ti - 16 sin 2tj} m>s2 dt a = 2a2 + a2 = 28.3232 + (-14.549)2 = 16.8 m>s2 x y When t = 1 s, a = -20 cos 2(1)i - 16 sin 2(1)j = {8.323i - 14.549j} m>s2. Thus, the magnitude of the acceleration is Ans. Traveling Path: Here, x = 5 cos 2t and y = 4 sin 2t. Then, x2 = cos2 2t 25 [1] y2 = sin2 2t 16 [2] Adding Eqs [1] and [2] yields y2 x2 + = cos2 2t + sin2 2t 25 16 However, cos2 2t + sin2 2t = 1. Thus, y2 x2 + = 1 25 16 (Equation of an Ellipse) (Q.E.D.) 63
64. 64. 91962_01_s12-p0001-0176 6/8/09 8:25 AM Page 64 © 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. 12–78. Pegs A and B are restricted to move in the elliptical slots due to the motion of the slotted link. If the link moves with a constant speed of 10 m/s, determine the magnitude of the velocity and acceleration of peg A when x = 1 m. y A C D x v ϭ 10 m/s Velocity: The x and y components of the peg’s velocity can be related by taking the first time derivative of the path’s equation. x2 + y2 = 1 4 1 # # (2xx) + 2yy = 0 4 1 # # xx + 2yy = 0 2 or 1 xv + 2yvy = 0 2 x At x = 1 m, (1)2 + y2 = 1 4 1 23 (1)(10) + 2 ¢ ≤ vy = 0 2 2 y = (1) 23 m 2 Here, vx = 10 m>s and x = 1. Substituting these values into Eq. (1), vy = -2.887 m>s = 2.887 m>s T v = 2vx 2 + vy 2 = 2102 + 2.8872 = 10.4 m>s Thus, the magnitude of the peg’s velocity is Ans. Acceleration: The x and y components of the peg’s acceleration can be related by taking the second time derivative of the path’s equation. 1 # # # # (xx + xx) + 2(yy + yy) = 0 2 1 #2 # A x + xx B + 2 A y 2 + yy B = 0 2 or 1 A v 2 + xax B + 2 A vy 2 + yay B = 0 2 x Since nx is constant, ax = 0. When x = 1 m, y = 23 m, vx = 10 m>s, and 2 (2) 23 1 a d = 0 A 102 + 0 B + 2 c(-2.887)2 + 2 2 y vy = -2.887 m>s. Substituting these values into Eq. (2), ay = -38.49 m>s2 = 38.49 m>s2 T a = 2ax 2 + ay 2 = 202 + (-38.49)2 = 38.5 m>s2 Thus, the magnitude of the peg’s acceleration is Ans. 64 B x2 ϩ y2 ϭ 1 4
65. 65. 91962_01_s12-p0001-0176 6/8/09 8:25 AM Page 65 © 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. 12–79. A particle travels along the path y2 = 4x with a constant speed of v = 4 m>s. Determine the x and y components of the particle’s velocity and acceleration when the particle is at x = 4 m. Velocity: The x and y components of the particle’s velocity can be related by taking the first time derivative of the path’s equation using the chain rule. # # 2yy = 4x 2 # # y = x y or At x = 4 m, y = 24(4) = 4 m. Thus Eq. (1) becomes vy = 2 v y x (1) vy = 1 v 2 x (2) v = 2vx 2 + vy 2 The magnitude of the particle’s velocity is (3) 2 1 vx 2 + a vx b A 2 Substituting v = 4 m>s and Eq. (2) into Eq. (3), 4 = Ans. vx = 3.578 m>s = 3.58 m>s Substituting the result of nx into Eq. (2), we obtain Ans. vy = 1.789 m>s = 1.79 m>s Acceleration: The x and y components of the particle’s acceleration can be related by taking the second time derivative of the path’s equation using the chain rule. # # 2(yy + yy) = 4x # y2 + yy = 2x or vy 2 + yay = 2ax (4) When x = 4 m, y = 4 m, and vy = 1.789 m>s. Thus Eq. (4) becomes 1.7892 + 4ay = 2ax ay = 0.5ax - 0.8 (5) Since the particle travels with a constant speed along the path, its acceleration along the tangent of the path is equal to zero. Here, the angle that the tangent makes with the dy 1 horizontal at x = 4 m is u = tan - 1 ¢ ≤ 2 = tan - 1 ¢ 1>2 ≤ 2 = tan - 1 (0.5) = 26.57°. dx x = 4 m x x=4 m Thus, from the diagram shown in Fig. a, ax cos 26.57° + ay sin 26.57° = 0 (6) Solving Eqs. (5) and (6) yields ax = 0.32 m>s2 ay = -0.64 m>s2 = 0.64 m>s2 T 65 Ans.
66. 66. 91962_01_s12-p0001-0176 6/8/09 8:26 AM Page 66 © 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. *12–80. The van travels over the hill described by y = ( -1.5(10–3) x2 + 15) ft. If it has a constant speed of 75 ft>s, determine the x and y components of the van’s velocity and acceleration when x = 50 ft. y 15 ft y ϭ (Ϫ1.5 (10Ϫ3) x2 ϩ 15) ft x 100 ft Velocity: The x and y components of the van’s velocity can be related by taking the first time derivative of the path’s equation using the chain rule. y = -1.5 A 10 - 3 B x2 + 15 # # y = -3 A 10 - 3 B xx or vy = -3 A 10 - 3 B xvx When x = 50 ft, vy = -3 A 10 - 3 B (50)vx = -0.15vx (1) v = 2vx 2 + vy 2 The magnitude of the van’s velocity is (2) 75 = 2vx 2 + (-0.15vx)2 Substituting v = 75 ft>s and Eq. (1) into Eq. (2), vx = 74.2 ft>s ; Ans. Substituting the result of nx into Eq. (1), we obtain vy = -0.15(-74.17) = 11.12 ft>s = 11.1 ft>s c Ans. Acceleration: The x and y components of the van’s acceleration can be related by taking the second time derivative of the path’s equation using the chain rule. \$ # # y = -3 A 10 - 3 B (xx + xx) or ay = -3 A 10 - 3 B A vx 2 + xax B When x = 50 ft, vx = -74.17 ft>s. Thus, ay = -3 A 10 - 3 B c(-74.17)2 + 50ax d ay = -(16.504 + 0.15ax) (3) Since the van travels with a constant speed along the path, its acceleration along the tangent of the path is equal to zero. Here, the angle that the tangent makes with the horizontal at dy x = 50 ft is u = tan - 1 ¢ ≤ 2 = tan - 1 c -3 A 10 - 3 B x d 2 = tan - 1( -0.15) = -8.531°. dx x = 50 ft x = 50 ft Thus, from the diagram shown in Fig. a, ax cos 8.531° - ay sin 8.531° = 0 (4) Solving Eqs. (3) and (4) yields ax = -2.42 ft>s = 2.42 ft>s2 ; Ans. ay = -16.1 ft>s = 16.1 ft>s2 T Ans. 66
67. 67. 91962_01_s12-p0001-0176 6/8/09 8:26 AM Page 67 © 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. y •12–81. A particle travels along the circular path from A to B in 1 s. If it takes 3 s for it to go from A to C, determine its average velocity when it goes from B to C. 30Њ Position: The coordinates for points B and C are [30 sin 45°, 30 - 30 cos 45°] and [30 sin 75°, 30 - 30 cos 75°]. Thus, C 45Њ 30 m rB = (30 sin 45° - 0)i + [(30 - 30 cos 45°) - 30]j B = {21.21i - 21.21j} m A rC = (30 sin 75° - 0)i + [(30 - 30 cos 75°) - 30]j = {28.98i - 7.765j} m Average Velocity: The displacement from point B to C is ¢rBC = rC - rB = (28.98i - 7.765j) - (21.21i - 21.21j) = {7.765i + 13.45j} m. (vBC)avg = 7.765i + 13.45j ¢rBC = = {3.88i + 6.72j} m>s ¢t 3 - 1 Ans. 12–82. A car travels east 2 km for 5 minutes, then north 3 km for 8 minutes, and then west 4 km for 10 minutes. Determine the total distance traveled and the magnitude of displacement of the car. Also, what is the magnitude of the average velocity and the average speed? Total Distance Traveled and Displacement: The total distance traveled is s = 2 + 3 + 4 = 9 km Ans. ¢r = 2(2 + 4)2 + 32 = 6.708 km = 6.71 km and the magnitude of the displacement is Ans. Average Velocity and Speed: The total time is ¢t = 5 + 8 + 10 = 23 min = 1380 s. The magnitude of average velocity is yavg = 6.708 A 103 B ¢r = = 4.86 m>s ¢t 1380 and the average speed is A ysp B avg Ans. 9 A 103 B s = = = 6.52 m>s ¢t 1380 Ans. 67 x
68. 68. 91962_01_s12-p0001-0176 6/8/09 8:27 AM Page 68 © 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. z 12–83. The roller coaster car travels down the helical path at constant speed such that the parametric equations that define its position are x = c sin kt, y = c cos kt, z = h - bt, where c, h, and b are constants. Determine the magnitudes of its velocity and acceleration. x = c sin kt # x = ck cos kt \$ x = -ck2 sin kt y = c cos kt # y = -ck sin kt \$ y = -ck2 cos kt z = h - bt # z = -b \$ z = 0 v = 2(ck cos kt)2 + (-ck sin kt)2 + (-b)2 = 2c2k2 + b2 y a = 2( -ck2 sin kt)2 + (-ck2 cos kt)2 + 0 = ck2 x Ans. Ans. *12–84. The path of a particle is defined by y2 = 4kx, and the component of velocity along the y axis is vy = ct, where both k and c are constants. Determine the x and y components of acceleration when y = y0. y2 = 4kx 2yvy = 4kvx 2v2 + 2yay = 4kax y vy = ct ay = c Ans. 2(ct)2 + 2yc = 4kax ax = c A y + ct2 B 2k Ans. 68
69. 69. 91962_01_s12-p0001-0176 6/8/09 8:27 AM Page 69 © 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. •12–85. A particle moves along the curve y = x - (x2>400), where x and y are in ft. If the velocity component in the x direction is vx = 2 ft>s and remains constant, determine the magnitudes of the velocity and acceleration when x = 20 ft. Velocity: Taking the first derivative of the path y = x - x2 , we have 400 1 # # # y = x (2xx) 400 x # # # y = x x 200 [1] # # However, x = yx and y = yy. Thus, Eq. [1] becomes yy = yx - x y 200 x [2] Here, yx = 2 ft>s at x = 20 ft. Then, From Eq. [2] yy = 2 - Also, 20 (2) = 1.80 ft>s 200 y = 2y2 + y2 = 222 + 1.802 = 2.69 ft>s x y Acceleration: Taking the second derivative of the path y = x - Ans. x2 , we have 400 1 #2 \$ \$ \$ y = x A x + xx B 200 [3] \$ \$ However, x = ax and y = ay. Thus, Eq. [3] becomes ay = ax - 1 A y2 + xax B 200 x [4] Since yx = 2 ft>s is constant, hence ax = 0 at x = 20 ft. Then, From Eq. [4] ay = 0 Also, 1 C 22 + 20(0) D = -0.020 ft>s2 200 a = 2a2 + a2 = 202 + ( -0.020)2 = 0.0200 ft>s2 x y Ans. 69
70. 70. 91962_01_s12-p0001-0176 6/8/09 8:28 AM Page 70 © 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. 12–86. The motorcycle travels with constant speed v0 along the path that, for a short distance, takes the form of a sine curve. Determine the x and y components of its velocity at any instant on the curve. y v0 π y ϭ c sin ( –– x) L x p y = c sin a xb L y = c c L L p p # cacos xbx L L vy = p p c vx acos xb L L v2 = v2 + v2 0 y x v2 = v2 B 1 + a 0 x vx = v0 B 1 + a vy = p 2 p cb cos2 a xb R L L -2 p 2 p cb cos2 a xb R L L 1 Ans. -2 v0 pc p p 2 p a cos xb B 1 + a cb cos2 a xb R L L L L 1 Ans. 12–87. The skateboard rider leaves the ramp at A with an initial velocity vA at a 30° angle. If he strikes the ground at B, determine vA and the time of flight. vA A Coordinate System: The x–y coordinate system will be set so that its origin coincides with point A. x-Motion: Here, (vA)x = vA cos 30°, xA = 0 and xB = 5 m. Thus, + A:B xB = xA + (vA)xt 5 = 0 + vA cos 30° t t = 5 vA cos 30° (1) y-Motion: Here, (vA)y = vA sin 30°, ay = -g = -9.81 m>s2, and yB = -1 m. Thus, A+cB yB = yA + (vA)y t + 1 a t2 2 y -1 = 0 + vA sin 30° t + 1 (-9.81)t2 2 4.905t2 - vA sin 30° t - 1 = 0 (2) Solving Eqs. (1) and (2) yields vA = 6.49 m>s 30Њ 1m t = 0.890 s Ans. 70 B 5m
71. 71. 91962_01_s12-p0001-0176 6/8/09 8:28 AM Page 71 © 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. *12–88. The pitcher throws the baseball horizontally with a speed of 140 ft>s from a height of 5 ft. If the batter is 60 ft away, determine the time for the ball to arrive at the batter and the height h at which it passes the batter. A + ; B 60 ft s = vt; 60 = 140t t = 0.4286 = 0.429 s A+cB 5 ft h s = s0 + v0 t + Ans. 1 a t2 2 c h = 5 + 0 + 1 (-32.2)(0.4286)2 = 2.04 ft 2 Ans. vA •12–89. The ball is thrown off the top of the building. If it strikes the ground at B in 3 s, determine the initial velocity vA and the inclination angle uA at which it was thrown. Also, find the magnitude of the ball’s velocity when it strikes the ground. uA A Coordinate System: The x–y coordinate system will be set so that its origin coincides with point A. 75 ft x-Motion: Here, (vA)x = vA cos u, xA = 0, and xB = 60 ft, and t = 3 s. Thus, + A:B xB = xA + (vA)xt 60 = 0 + vA cos u(3) vA cos u = 20 (1) y-Motion: Here, (vA)y = vA sin u, ay = -g = -32.2 ft>s , yA = 0, and yB = -75 ft, and t = 3 s. Thus, A+cB 1 2 a t 2 y 1 -75 = 0 + vA sin u(3) + (-32.2) A 32 B 2 yB = yA + (vA)y t + vA sin u = 23.3 (2) Solving Eqs. (1) and (2) yields u = 49.36° = 49.4° vA = 30.71 ft>s = 30.7 ft>s Ans. Using the result of u and nA, we obtain (vA)x = 30.71 cos 49.36° = 20 ft>s (vA)y = 30.71 sin 49.36° = 23.3 ft>s Thus, A+cB B 60 ft 2 (vB)y = (vA)y + ayt (vB)y = 23.3 + (-32.2)(3) = -73.3 ft>s = 73.3 ft>s T vB = 2202 + 73.32 = 76.0 ft>s Thus, the magnitude of the ball’s velocity when it strikes the ground is Ans. 71
72. 72. 91962_01_s12-p0001-0176 6/8/09 8:29 AM Page 72 © 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. y 12–90. A projectile is fired with a speed of v = 60 m>s at an angle of 60°. A second projectile is then fired with the same speed 0.5 s later. Determine the angle u of the second projectile so that the two projectiles collide. At what position (x, y) will this happen? v ϭ 60 m/s 60Њ u v ϭ 60 m/s x x x-Motion: For the motion of the first projectile, vx = 60 cos 60° = 30 m>s, x0 = 0, and t = t1. Thus, + A:B x = x0 + vxt x = 0 + 30t1 (1) For the motion of the second projectile, vx = 60 cos u, x0 = 0, and t = t1 - 0.5. Thus, + A:B x = x0 + vxt x = 0 + 60 cos u(t1 - 0.5) (2) y-Motion: For the motion of the first projectile, vy = 60 sin 60° = 51.96 m>s, y0 = 0, and ay = -g = -9.81 m>s2. Thus, A+cB 1 2 at 2 y 1 y = 0 + 51.96t1 + (-9.81)t1 2 2 y = y0 + vyt + y = 51.96t1 - 4.905t1 2 For the motion of the ay = -g = -9.81 m>s2. Thus, A+cB y = y0 + vyt + second (3) projectile, vy = 60 sin u, y0 = 0, and 1 2 at 2 y y = 0 + 60 sin u(t1 - 0.5) + 1 ( -9.81)(t1 - 0.5)2 2 y = (60 sin u)t1 - 30 sin u - 4.905 t1 2 + 4.905t1 - 1.22625 (4) Equating Eqs. (1) and (2), 30t1 = 60 cos u(t1 - 0.5) t1 = cos u 2 cos u - 1 (5) Equating Eqs. (3) and (4), 51.96t1 - 4.905t1 2 = (60 sin u)t1 - 30 sin u - 4.905t1 2 + 4.905t1 - 1.22625 (60 sin u - 47.06)t1 = 30 sin u + 1.22625 t1 = 30 sin u + 1.22625 60 sin u - 47.06 (6) 72 y
73. 73. 91962_01_s12-p0001-0176 6/8/09 8:29 AM Page 73 © 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. Equating Eqs. (5) and (6) yields cos u 30 sin u + 1.22625 = 2 cos u - 1 60 sin u - 47.06 49.51 cos u - 30 sin u = 1.22625 Solving by trial and error, Ans. u = 57.57° = 57.6° Substituting this result into Eq. (5) (or Eq. (6)), t1 = cos 57.57° = 7.3998 s 2 cos 57.57° - 1 Substituting this result into Eqs. (1) and (3), x = 30(7.3998) = 222 m Ans. y = 51.96(7.3998) - 4.905 A 7.39982 B = 116 m Ans. 73
74. 74. 91962_01_s12-p0001-0176 6/8/09 8:30 AM Page 74 © 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. 12–91. The fireman holds the hose at an angle u = 30° with horizontal, and the water is discharged from the hose at A with a speed of vA = 40 ft>s. If the water stream strikes the building at B, determine his two possible distances s from the building. vA ϭ 40 ft/s A B u 8 ft 4 ft Coordinate System: The x–y coordinate system will be set so that its origin coincides with point A. x-Motion: Here, (vA)x = 40 cos 30° ft>s = 34.64 ft>s, xA = 0, and xB = s. Thus, + A:B xB = xA + (vA)xt s = 0 + 34.64t s = 34.64t (1) y-Motion: Here, (vA)y = 40 sin 30° ft>s = 20 ft>s, ay = -g = -32.2 ft>s2, yA = 0, and yB = 8 - 4 = 4 ft. Thus, A+cB yB = yA + (vA)y t + 4 = 0 + 20t + 1 a t2 2 y 1 (-32.2)t2 2 16.1t2 - 20t + 4 = 0 t = 0.2505 s and 0.9917 s Substituting these results into Eq. (1), the two possible distances are s = 34.64(0.2505) = 8.68 ft Ans. s = 34.64(0.9917) = 34.4 ft Ans. 74 s
75. 75. 91962_01_s12-p0001-0176 6/8/09 8:30 AM Page 75 © 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. *12–92. Water is discharged from the hose with a speed of 40 ft>s. Determine the two possible angles u the fireman can hold the hose so that the water strikes the building at B. Take s = 20 ft. vA ϭ 40 ft/s A B u 8 ft 4 ft Coordinate System: The x–y coordinate system will be set so that its origin coincides with point A. x-Motion: Here, (vA)x = 40 cos u, xA = 0, and xB = 20 ft>s. Thus, + A:B xB = xA + (vA)xt 20 = 0 + 40 cos ut t = 1 2 cos u (1) y-Motion: Here, (vA)y = 40 sin u, ay = -g = -32.2 ft>s2, yA = 0, and yB = 8 - 4 = 4 ft. Thus, A+cB 1 a t2 2 y 1 4 = 0 + 40 sin ut + (-32.2)t2 2 yB = yA + (vA)yt + 16.1t2 - 40 sin ut + 4 = 0 (2) Substituting Eq. (1) into Eq. (2) yields 16.1 ¢ 2 1 1 ≤ - 40 sin u ¢ ≤ + 4 = 0 2 cos u 2 cos u 20 sin u cos u - 4 cos2 u = 4.025 10 sin u cos u - 2 cos2 u = 2.0125 5 sin 2u - A 2 cos2 u - 1 B - 1 = 2.0125 5 sin 2u - cos 2u = 3.0125 Solving by trial and error, u = 23.8° and 77.5° Ans. 75 s