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: 47

Answer

$s(t)=e^t+19t+4$

Work Step by Step

Here, $a=e^t; v(0)=20;s(0)=5$ Since the derivative of velocity is acceleration, $v(t)=e^t+C$ Therefore, $v(0)=20$ Then $20=e^0+C \implies C=19$ Thus, $v(t)=e^t+19$ Since, the derivative of position (that is, $s(t)$) is velocity, $s(t)=e^t+19t+C'$ Therefore, $s(0)=5$ Then $5=e^0+19(0)+C' \implies C'=4$ Hence, $s(t)=e^t+19t+4$
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