Answer
$a=7.42$ ft/s$^2$
Work Step by Step
$\int_{15}^v dv = \int_0^5 0.8t dt$
$v=25$ ft/s
$a_n=\frac{v^2}{\rho}=\frac{25^2}{100}=6.25$ ft/s$^2$
At $t=5$ s,
$a_t=\dot{v}=0.8(5)=4$ ft/s$^2$
$a=\sqrt{a_t^2+a_n^2}=\sqrt{4^2+6.25^2}=7.42$ 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 12.7 - Curvilinear Motion: Normal and Tangential Components - Problems - Page 68: 130
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