Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.8 - Modeling Using Variation - Exercise Set - Page 488: 26

Answer

$120\; feet$.

Work Step by Step

Step 1:- Translate the statement to form an equation. Let the distance be $D$. and the speed be $S$. $\Rightarrow D=kS^2$ ...... (1) Step 2:- Substitute the first set of values into the equation to find the value of $k$. The given values are $S=45\; miles\; per\; hour$ and $D=67.5\; feet$. Substitute into the equation (1). $\Rightarrow 67.5=k(45)^2$ $\Rightarrow 67.5=2025k$ Divide both sides by $2025$. $\Rightarrow \frac{67.5}{2025}=\frac{2025k}{2025}$ Simplify. $\Rightarrow \frac{1}{30}=k$ Step 3:- Substitute the value of $k$ into the original equation. Substitute $k=\frac{1}{30}$ into the equation (1). $\Rightarrow D=\frac{1}{30}S^2$ ...... (2) Step 4:- Solve the equation to find the required value. Substitute $S=60\; miles\; per\; hour$ into the equation (2). $\Rightarrow D=\frac{1}{30}(60)^2$ Simplify. $\Rightarrow D=120$ Hence, the stopping distance is $120\; feet$.
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