Answer
$s=123$ ft
$a=2.99$ ft/s$^2$
Work Step by Step
$v=60(1-e^{-t})$
$\int_0^s ds = \int v dt = \int_0^3 60(1-e^{-t})dt$
$s=60(t+e^{-t})|^3_0$
$s=123$ ft
$a=\frac{dv}{dt}=60(e^{-t})$
At $t=3$ s
$a=60e^{-3}=2.99$ ft/s$^2$
Engineering Mechanics: Statics & Dynamics (14th Edition)
by Hibbeler, Russell C.
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: 21
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