Q. 89 - Precalculus Enhanced with Graphing Utilities

Question

Vehicle Stopping Distance Taking into account reaction time, the distance $d$ (in feet) that a car requires to come to a complete stop while traveling $r$ miles per hour is given by the function

$d\left(r\right)=6.97 r-90.39$

a. Express the speed $r$ at which the car is traveling as a function of the distance $d$ required to come to a complete stop

b. Verify that $r=r\left(d\right)$ is the inverse of $d=d\left(r\right)$ by showing that $r\left(d\right(r\left)\right)=r$ and $d\left(r\right(d\left)\right)=d$

c. Predict the speed that a car was traveling if the distance required to stop was $300$ feet.

$d\left(r\right(d\left)\right)=d$

Step-by-Step Solution

Answer

The values are $r\left(d\right)=\frac{d+90.39}{6.97}$, $56$ respectively

Part (a) Step 1: Given information

Given the function $d\left(r\right)=6.97 r-90.39$

Part (b) Step 2: Replace d ( r ) by d

Replacing, we get

$$\begin{gathered} d=6.97 r-90.39 \\ d+90.39=6.97 r \\ \frac{d+90.39}{6.97}=\frac{6.97r}{6.97} \\ \frac{d+90.39}{6.97}=r \\ r=\frac{d+90.39}{6.97} \\ r\left(d\right)=\frac{d+90.39}{6.97} \end{gathered}$$

Part (b) Step 1: Given information

Given the function $d\left(r\right)=6.97 r-90.39$

Part (b) Step 2: Verify the function

Verifying, we get

$$\begin{gathered} r\left(d\right(r\left)\right)=r(6.97 r-90.39) \\ r\left(d\right(r\left)\right)=\frac{6.97r-90.39+90.39}{6.97} \\ r\left(d\right(r\left)\right)=r \end{gathered}$$

$$\begin{gathered} d\left(r\right(d\left)\right)=6.97\left(\frac{d+90.39}{6.97}\right)-90.39 \\ d\left(r\right(d\left)\right)=d+90.39-90.39 \\ d\left(r\right(d)=d \end{gathered}$$

Part (c) Step 1: Given information

Given the function $d\left(r\right)=6.97 r-90.39$

Part (c) Step 2: Substituting and calculating the value

Substituting, we get

$$\begin{gathered} r\left(d\right)=\frac{d+90.39}{6.97} \\ r\left(300\right)=\frac{300+90.39}{6.97} \\ r\left(300\right)=\frac{390.39}{6.97} \\ r\left(300\right)\approx 56 \end{gathered}$$