University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 224: 49


$s(t)= \sin 2t-3$

Work Step by Step

Since the derivative of velocity is acceleration, $v(t)=2 \cos 2t+C$ As we have, $v(0)=2$ Then $2=2 \cos 2(0)+C \implies C=0$ Thus, $v(t)=2 \cos 2t$ Since the derivative of position is velocity, $s(t)=\sin 2t+C'$ Therefore, $s(0)=-3$ Then $-3=\sin 2(0)+C' \implies C'=-3$ Hence, $s(t)= \sin 2t-3$
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