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 18: 14

Answer

v= 165 ft/s $a= 48ft/s^{2}$ $s_t= 450 ft$ $v_avg= 25 ft/s$ $(v_sp)v_avg=45 ft/s$

Work Step by Step

$s=t^3-6t^2-15t+7$ $v=\frac{ds}{dt}=3t^2-12t-15$ Substituting t when t is 10s v= 165ft/s $a=\frac{dv}{dt}=6t-12$ therefore when t is 10s $\:a=\frac{48ft}{^{s^2}}$ If v = 0 $0=3t^2-12t-15$ Evaluating the equation above, we get its positive root to be t = 5s Therefore: If t = 0. s = 7ft If t=5s, s = -93 ft If t = 10s, s= 257ft The total distance will be: $s_t = 7+ 93+93+257 = 450 ft$ instantaneous velocity velocity $(v_sp)v_avg=\frac{ds}{dt}=\frac{257-7}{10-0}= 25ft/s$ Instantaneous acceleration $(v_sp)_avg=$\frac{s_t}{dt}=\frac{450}{10}=45 ft/s$
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