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 $LaTeX: \displaystyle 63$ miles per hour can stop in $LaTeX: \displaystyle 161$ 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 $LaTeX: \displaystyle 161=k(63)^2 \iff k=\frac{23}{567}$ . The model is $LaTeX: \displaystyle d=\frac{23}{567}r$ evaluating at $LaTeX: \displaystyle r=36$ gives $LaTeX: \displaystyle d=\frac{368}{7}\approx 53$ feet