University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 147: 31


The time it takes the aircraft to be airborne is $25\ sec$. It has traveled $694.4\ m$ during that time.

Work Step by Step

$$D=\frac{10}{9}t^2(m)$$ The velocity of the aircraft is the derivative of the distance it has traveled, $D$: $$v=\frac{dD}{dt}=\frac{20}{9}t (m/s)$$ The speed of the aircraft, thus, is $|v|=|20/9t|=20/9 t$ (m/s) (because the time $t$ is positive). Convert $200km/h$ to m/s: $$200km/h=\frac{200\times1000}{3600}m/s=\frac{500}{9}m/s$$ The time it takes the aircraft to reach $200km/h$ is $$t=\frac{\frac{500}{9}}{\frac{20}{9}}=25(sec)$$ The distance it has traveled after $25sec$ is $$D=\frac{10}{9}\times(25^2)\approx694.4(m)$$
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