College Algebra, Chapter 2, 2.5, Section 2.5, Problem 36

Suppose that the stopping distance $D$ of a car after the breaches have been applied varies directly as the square of the speed $s$. A certain car travelling at $\displaystyle 50 \frac{\text{mi}}{\text{h}}$ can stop in 240 ft. What is the maximum speed it can be travelling if it needs to stop in 160ft?

$
\begin{equation}
\begin{aligned}
D &= ks ^2 && \text{Model}\\
\\
240 \text{ft} \times \frac{1 \text{mile}}{5280 \text{ft}} &= k \left( 50 \frac{\text{mi}}{\text{h}} \right)^2 && \text{Substitute the given and convert ft into miles}\\
\\
\frac{1}{22} \text{mi} &= k \left(2500 \frac{\text{mi}^2}{\text{h}^2} \right) && \text{Simplify}\\
\\
k &= \frac{\frac{1}{22}\text{mi}}{2500 \frac{\text{mi}^2}{\text{h}^2}}\\
\\
k &= \frac{1}{55000} \frac{\text{h}^2}{\text{mi}}
\end{aligned}
\end{equation}
$


Then if,

$
\begin{equation}
\begin{aligned}
D &= 160 \text{ft},\\
\\
D &= ks^2\\
\\
160 \text{ft} \times \frac{1 \text{mile}}{5280 \text{ft}} &= \frac{1}{55000} \frac{\text{h}^2}{\text{mi}} (s^2) && \text{Solve for }s \\
\\
\frac{1}{33} \text{mi} &= \frac{1}{55000} \frac{\text{h}^2}{\text{mi}} (s^2)\\
\\
s^2 &= \frac{55000}{33} \frac{\text{mi}^2}{\text{h}^2} && \text{Take the square root}\\
\\
s &= \frac{50\sqrt{6}}{3} \frac{\text{mi}}{\text{h}}
\end{aligned}
\end{equation}
$


It shows that if the car needs to stop in 160ft. Its speed must be a maximum of $\displaystyle \frac{50\sqrt{6}}{3} \frac{\text{mi}}{\text{h}}$