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 223: 43



Work Step by Step

Here, $s=\dfrac{ds}{dt}=9.8t+5$, thus the general form of the function is: $s(t)=4.9t^2+5t+C$ Since we have $s(0)=10$ Therefore, $s(0)=10 \implies 10=4.9(0)^2+5(0)+C$ Thus, $C=10$ Hence, $s(t)=4.9t^2+5t+10$
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