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The breaking distance, \(\displaystyle d\), of a car is proportional to the square of the speed, \(\displaystyle r\). If a car traveling at \(\displaystyle 65\) miles per hour can stop in \(\displaystyle 103\) 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 \(\displaystyle d=kr^2\). Evaluating at the known values and solving for \(\displaystyle k\) gives \(\displaystyle 103=k(65)^2 \iff k=\frac{103}{4225}\). The model is \(\displaystyle d=\frac{103}{4225}r\) evaluating at \(\displaystyle r=36\) gives \(\displaystyle d=\frac{133488}{4225}\approx 32\) feet

Download \(\LaTeX\) 

\begin{question}The breaking distance, $d$, of a car is proportional to the square of the speed, $r$. If a car traveling at $65$ miles per hour can stop in $103$ feet. What is the stopping distance of a car traveling 36 miles per hour? Round the solution to the nearest foot.
    \soln{9cm}{The equation of variation is $d=kr^2$.  Evaluating at the known values and solving for $k$ gives $103=k(65)^2 \iff k=\frac{103}{4225}$. The model is $d=\frac{103}{4225}r$ evaluating at $r=36$ gives $d=\frac{133488}{4225}\approx 32$ feet}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>The breaking distance,  <img class="equation_image" title=" \displaystyle d " src="/equation_images/%20%5Cdisplaystyle%20d%20" alt="LaTeX:  \displaystyle d " data-equation-content=" \displaystyle d " /> , of a car is proportional to the square of the speed,  <img class="equation_image" title=" \displaystyle r " src="/equation_images/%20%5Cdisplaystyle%20r%20" alt="LaTeX:  \displaystyle r " data-equation-content=" \displaystyle r " /> . If a car traveling at  <img class="equation_image" title=" \displaystyle 65 " src="/equation_images/%20%5Cdisplaystyle%2065%20" alt="LaTeX:  \displaystyle 65 " data-equation-content=" \displaystyle 65 " />  miles per hour can stop in  <img class="equation_image" title=" \displaystyle 103 " src="/equation_images/%20%5Cdisplaystyle%20103%20" alt="LaTeX:  \displaystyle 103 " data-equation-content=" \displaystyle 103 " />  feet. What is the stopping distance of a car traveling 36 miles per hour? Round the solution to the nearest foot.</p> </p>
HTML for Canvas
<p> <p>The equation of variation is  <img class="equation_image" title=" \displaystyle d=kr^2 " src="/equation_images/%20%5Cdisplaystyle%20d%3Dkr%5E2%20" alt="LaTeX:  \displaystyle d=kr^2 " data-equation-content=" \displaystyle d=kr^2 " /> .  Evaluating at the known values and solving for  <img class="equation_image" title=" \displaystyle k " src="/equation_images/%20%5Cdisplaystyle%20k%20" alt="LaTeX:  \displaystyle k " data-equation-content=" \displaystyle k " />  gives  <img class="equation_image" title=" \displaystyle 103=k(65)^2 \iff k=\frac{103}{4225} " src="/equation_images/%20%5Cdisplaystyle%20103%3Dk%2865%29%5E2%20%5Ciff%20k%3D%5Cfrac%7B103%7D%7B4225%7D%20" alt="LaTeX:  \displaystyle 103=k(65)^2 \iff k=\frac{103}{4225} " data-equation-content=" \displaystyle 103=k(65)^2 \iff k=\frac{103}{4225} " /> . The model is  <img class="equation_image" title=" \displaystyle d=\frac{103}{4225}r " src="/equation_images/%20%5Cdisplaystyle%20d%3D%5Cfrac%7B103%7D%7B4225%7Dr%20" alt="LaTeX:  \displaystyle d=\frac{103}{4225}r " data-equation-content=" \displaystyle d=\frac{103}{4225}r " />  evaluating at  <img class="equation_image" title=" \displaystyle r=36 " src="/equation_images/%20%5Cdisplaystyle%20r%3D36%20" alt="LaTeX:  \displaystyle r=36 " data-equation-content=" \displaystyle r=36 " />  gives  <img class="equation_image" title=" \displaystyle d=\frac{133488}{4225}\approx 32 " src="/equation_images/%20%5Cdisplaystyle%20d%3D%5Cfrac%7B133488%7D%7B4225%7D%5Capprox%2032%20" alt="LaTeX:  \displaystyle d=\frac{133488}{4225}\approx 32 " data-equation-content=" \displaystyle d=\frac{133488}{4225}\approx 32 " />  feet</p> </p>