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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 \(\displaystyle 59\) miles per hour can stop in \(\displaystyle 182\) 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 \(\displaystyle 182=k(59)^2 \iff k=\frac{182}{3481}\). The model is \(\displaystyle d=\frac{182}{3481}r\) evaluating at \(\displaystyle r=36\) gives \(\displaystyle d=\frac{235872}{3481}\approx 68\) feet
\begin{question}The breaking distance, $d$, of a car is proportional to the square of the speed, $r$. If a car traveling at $59$ miles per hour can stop in $182$ 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 $182=k(59)^2 \iff k=\frac{182}{3481}$. The model is $d=\frac{182}{3481}r$ evaluating at $r=36$ gives $d=\frac{235872}{3481}\approx 68$ feet} \end{question}
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<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 59 " src="/equation_images/%20%5Cdisplaystyle%2059%20" alt="LaTeX: \displaystyle 59 " data-equation-content=" \displaystyle 59 " /> miles per hour can stop in <img class="equation_image" title=" \displaystyle 182 " src="/equation_images/%20%5Cdisplaystyle%20182%20" alt="LaTeX: \displaystyle 182 " data-equation-content=" \displaystyle 182 " /> feet. What is the stopping distance of a car traveling 36 miles per hour? Round the solution to the nearest foot.</p> </p>
<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 182=k(59)^2 \iff k=\frac{182}{3481} " src="/equation_images/%20%5Cdisplaystyle%20182%3Dk%2859%29%5E2%20%5Ciff%20k%3D%5Cfrac%7B182%7D%7B3481%7D%20" alt="LaTeX: \displaystyle 182=k(59)^2 \iff k=\frac{182}{3481} " data-equation-content=" \displaystyle 182=k(59)^2 \iff k=\frac{182}{3481} " /> . The model is <img class="equation_image" title=" \displaystyle d=\frac{182}{3481}r " src="/equation_images/%20%5Cdisplaystyle%20d%3D%5Cfrac%7B182%7D%7B3481%7Dr%20" alt="LaTeX: \displaystyle d=\frac{182}{3481}r " data-equation-content=" \displaystyle d=\frac{182}{3481}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{235872}{3481}\approx 68 " src="/equation_images/%20%5Cdisplaystyle%20d%3D%5Cfrac%7B235872%7D%7B3481%7D%5Capprox%2068%20" alt="LaTeX: \displaystyle d=\frac{235872}{3481}\approx 68 " data-equation-content=" \displaystyle d=\frac{235872}{3481}\approx 68 " /> feet</p> </p>