Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 241: 102

Answer

$1.24sec$

Work Step by Step

Step 1. Given $\frac{d^2s}{dt^2}=-5.2ft/s^2$, we can get $\frac{ds}{dt}=\int(-5.2)dt=-5.2t+C$ and $s=\int(-5.2t+C)dt=-2.6t^2+Ct+D$, where $C$ and $D$ are constants. Step 2. At $t=0, \frac{ds}{dt}=0$, we have $-5.2(0)+C=0$ and $C=0$ Step 3. At $t=0, s=4$, we have $-2.6(0)^2+D=4$ and $D=4$. Thus $s=\int(-5.2t+C)dt=-2.6t^2+4$ Step 4. Let $s(t)=0$, we have $-2.6t^2+4=0$ which gives $t=\sqrt {\frac{20}{13}}\approx1.24sec$ Step 5. It takes the hammer and feather about $1.24$ seconds to fall 4 ft on the moon.
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