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: 77

Answer

A constant deceleration of $~~-4.82~m/s^2~~$ is required.

Work Step by Step

We can express $100~km/h$ in units of $m/s$: $(100~km/h)\times \frac{1000~m}{1~km}\times \frac{1~h}{3600~s} = 27.78~m/s$ $a(t) = \frac{dv}{dt} = a$ $v(t) = v_0+a~t$ $s(t) = s_0+v_0~t+\frac{1}{2}at^2$ When $t = 0$, we can let $s = 0$. Then $s_0 = 0$ $s(t) = v_0~t+\frac{1}{2}at^2$ When $s = 80$, we can let $v = 0$ Then $at = -v_0$ We can find $t$ when $s = 80$: $s(t) = v_0~t+\frac{1}{2}at^2 = 80$ $v_0~t+\frac{1}{2}(-v_0)~t = 80$ $\frac{1}{2}(v_0)~t = 80$ $t = \frac{160}{v_0}$ $t = \frac{160}{27.78}$ $t = 5.76~s$ We can find $a$: $v(5.76) = v_0+a~t = 0$ $a~t = -v_0$ $a = -\frac{v_0}{t}$ $a = -\frac{27.78}{5.76}$ $a = -4.82~m/s^2$ A constant deceleration of $~~-4.82~m/s^2~~$ is required.
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