Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.9 Antiderivatives - 3.9 Exercises - Page 284: 69

Answer

$14400 \text{ miles per squared hour}$

Work Step by Step

The acceleration $\text{a}$ is the instantaneous rate of change of the speed in a fixed small period of time: $$a=\frac{d v}{d t} \to dv=a dt \to \int dv=\int a dt \to \int dv=a\int dt $$ $$\to \Delta v=a \Delta t \to a=\frac{\Delta v}{\Delta t} $$ To increase, the variation in speed $ \Delta v$ should be positive. Notice that $5$ seconds is equivalent to $\frac{5}{3600}$ hours. $$a=\frac{\Delta v}{\Delta t} \to a=\frac{50-30}{\frac{5}{3600}}=14400 \text{ miles per squared hour} $$
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