Braking distance How is the braking distance for a motorcycle related to the speed at which the motorcycle was traveling when the brake was applied? Statistics teacher Aaron Waggoner gathered data to answer this question. The table shows the speed (in miles per hour) and the distance needed to come to a complete stop when the brake was applied (in feet).

$\begin{array}{|llll|}\hline \text{Speed (mph)}& \text{Distance (ft)}& \text{Speed (mph)}& \text{Distance (ft)}\\ 6& 1.42& 32& 52.08\\ 9& 4.92& 40& 84\\ 19& 18& 48& 110.33\\ 30& 44.75& & \\ \hline\end{array}$

a. Transform both variables using logarithms. Then calculate and state the least-squares regression line using the transformed variables.

b. Use the model from part (a) to calculate and interpret the residual for the trial when the motorcycle was traveling at $48$ mph.

Given data:

$\begin{array}{|llll|}\hline \text{Speed (mph)}& \text{Distance (ft)}& \text{Speed (mph)}& \text{Distance (ft)}\\ 6& 1.42& 32& 52.08\\ 9& 4.92& 40& 84\\ 19& 18& 48& 110.33\\ 30& 44.75& & \\ \hline\end{array}$

Log of given data:$\begin{array}{|llll|}\hline \text{Speed (mph)}& \text{Distance (ft)}& \log (\text{speed)}& \log (\text{distance)}\\ 6& 1.42& 0.7781513& 0.15228834\\ 9& 4.92& 0.9542425& 0.6919651\\ 19& 18& 1.2787536& 1.25527251\\ 30& 44.75& 1.4771213& 1.65079304\\ 32& 52.08& 1.50515& 1.71667098\\ 40& 84& 1.60206& 1.92427929\\ 48& 110.33& 1.6812412& 2.04269362\\ \hline\end{array}$

Making use of a Ti83/84 calculator

Step 1: Press STAT,

Step 2: select 1: EDIT.

Step 3: in list $L1$, type the data for logarithmic speed, and in list $L2$, type the data for logarithmic distance.

Step 4: hit STAT once more, select CALC, and then $LinReg(a+bx)$.

The result:

$y=a+bx$

$a=-1.3506$

$b=2.0361$

Substituting the value in $a$and $b$

$y=-1.3506+2.0361x$

The logarithm of speed is $x$, whereas the logarithm of distance is $y$.

localid="1654264833872" $\log y=-1.3506+2.0361\log x$

$x$denotes the speed and $y$denotes the distance.

Given data:

$\begin{array}{|llll|}\hline \text{Speed (mph)}& \text{Distance (ft)}& \text{Speed (mph)}& \text{Distance (ft)}\\ 6& 1.42& 32& 52.08\\ 9& 4.92& 40& 84\\ 19& 18& 48& 110.33\\ 30& 44.75& & \\ \hline\end{array}$

Substituting the value $x$by $48$:

$\log y=-1.3506+2.0361\log x$

$\log y=-1.3506+2.0361\log 48$

$\log y=2.0726$

Using the exponential function with a base of ten:

$y={10}^{\log y}$

$={10}^{2.0726}$

$=118.195$