## Related questions with answers

The brakes of a certain automobile produce a constant deceleration of $k \ \mathrm{ft} / \mathrm{s}^{2}$. 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?

Solution

VerifiedLet $a(t), v(t), s(t)$ denote the acceleration, velocity and position of the car at $t$ seconds. Given that $a(t)=-K$, here negative sign indicates that the car is decelerating and $K$ is a constant. Then

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

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

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

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

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