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Question

The brakes of a certain automobile produce a constant deceleration of k ft/s2k \ \mathrm{ft} / \mathrm{s}^{2}. The car is traveling at 60 mi/h (88 ft/s) when the driver is forced to hit the brakes, and it comes to rest at a point 121 ft from the point where the brakes were applied. What is k?

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Answered 2 years ago
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Let a(t),v(t),s(t)a(t), v(t), s(t) denote the acceleration, velocity and position of the car at tt seconds. Given that a(t)=Ka(t)=-K, here negative sign indicates that the car is decelerating and KK is a constant. Then

v(t)=a(t)dt=Kdt=Kt+C1\begin{aligned}v(t)&=\int{a(t)\:dt}\\&=\int{-K\:dt}\\&=-Kt+C_1\end{aligned}

Now since the initial velocity of the car is 88 ft/s88\text{ ft/s}, that is v(0)=88 ft/sv(0)=88\text{ ft/s}, thus we get

v(0)=8888=K(0)+C1C1=88\begin{aligned}v(0)&=88\\ 88&=K(0)+C_1\\ C_1&=88\end{aligned}

Tis follows that v(t)=Kt+88v(t)=-Kt+88.

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