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 $\displaystyle \frac{50\sqrt{6}}{3} \frac{\text{mi}}{\text{h}}$