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: 12


$t=30$s $s=792$m

Work Step by Step

We solve for time as follows: $v=v_1+a_ct$ $120=70+6000t$ $t=8.33\times10^{-3}$ hr = $30$ s Next, we solve for distance: $v^2=v_1^2+2a_c(s-s_1)$ $120^2=70^2+2(6000)(s-0)$ $s=0.792$ km $= 792$ m
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