Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises - Page 357: 66

Answer

$s=\frac{1}{2}at^2+v_0t+s_0$ (see work)

Work Step by Step

Note that $s''=a$. Since $a$ is constant, $a(t)=a.$ Then $v'=a$, so $v=at+C.$ But we also know that $v(0)=v_0,$ hence $v=at+v_0.$ Then $s'=v$, so $s=\frac{1}{2}at^2+v_0t+D$, and again we know that $s(0)=s_0$, so it follows that $s=\frac{1}{2}at^2+v_0t+s_0$.
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