Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 12 - Kinematics of a Particle - Section 12.2 - Rectilinear Kinematics: Continuous Motion - Problems - Page 17: 4


$v=\sqrt{9.17s-27.7}$ ft/s

Work Step by Step

We solve for the acceleration as follows: $v^2=v_0^2+2a_c(s-s_0)$ $8^2=3^2+2a_c(10-4)$ $a_c=4.583$ ft/s$^2$ Next, we solve for velocity: $v^2=v_0^2+2a_c(s-s_0)$ $v^2=3^2+2(4.583)(s-4)$ $v=\sqrt{9.17s-27.7}$ ft/s
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