The breaking distance, $LaTeX: \displaystyle d$ , of a car is proportional to the square of the speed, $LaTeX: \displaystyle r$ . If a car traveling at $LaTeX: \displaystyle 55$ miles per hour can stop in $LaTeX: \displaystyle 237$ feet. What is the stopping distance of a car traveling 36 miles per hour? Round the solution to the nearest foot.

The equation of variation is $LaTeX: \displaystyle d=kr^2$ . Evaluating at the known values and solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle 237=k(55)^2 \iff k=\frac{237}{3025}$ . The model is $LaTeX: \displaystyle d=\frac{237}{3025}r$ evaluating at $LaTeX: \displaystyle r=36$ gives $LaTeX: \displaystyle d=\frac{307152}{3025}\approx 102$ feet