In his revision of Engineering Mechanics, R.C. Hibbeler empowers scholars to achieve the total studying event. Hibbeler achieves this by means of calling on his daily school room adventure and his wisdom of ways scholars examine in and out of lecture.
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Additional resources for Engineering Mechanics: Dynamics, Instructor Solutions Manual
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. SOLUTION 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 v0 = 0. 25(t¿)2 2 The velocity at time t can be obtained by applying Eq. 12–4 with v0 = 0. 5t¿ (Eq. 5(t¿)2 in + B A: permitted. Dissemination Wide copyright If the total distance traveled is sTot = 2000, then is of the not instructors States World sTot = s1 + s2 by and use United on learning.
Construct the v- s graph for 0 … s … s¿ . a(ft/s2) 6 1000 SOLUTION v 4 s Graph: For 0 … s 6 1000 ft, the initial condition is v = 0 at s = 0. 54 ft>s = 110 ft>s Dissemination World permitted. 54 ft>s at s = 1000 ft. v is of and use United on learning. 54 ft>s this assessing A 220 000 - 8s B ft>s part the is and any courses When v = 0, This provided integrity of and work v = s¿ = 2500 ft The v–s graph is shown in Fig. a. , Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12–37), 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. 12–37. SOLUTION From Prob. 12–37, (+ c ) a = -g0 R2 (R + y)2 Since a dy = v dv then v = L0 v dv y 1 v2 d = R + y y0 2 g0 R2[ 1 1 v2 ] = R + y R + y0 2 Web) laws or g0 R2 c teaching Dissemination Wide Ly0 (R + y) 2 in dy copyright y -g0 R2 States World permitted.
Engineering Mechanics: Dynamics, Instructor Solutions Manual by HIBBELER