The stopping distance of an automobile, on dry, level pavement, traveling at a speed $v$ (kilometers per hour) is the distance $R$ (meters) the car travels during the reaction time of the driver plus the distance $B$ (meters) the car travels after the brakes are applied (see figure). The table shows the results of an experiment.
$$\begin{array}{|l|c|c|c|c|c|}\hline \text { Speed, } \boldsymbol{v} & 20 & 40 & 60 & 80 & 100 \\\hline \begin{array}{l}\text { Reaction Time } \\\text { Distance, } \boldsymbol{R}\end{array} & 8.3 & 16.7 & 25.0 & 33.3 & 41.7 \\\hline \begin{array}{l}\text { Braking Time } \\\text { Distance, } \boldsymbol{B}\end{array} & 2.3 & 9.0 & 20.2 & 35.8 & 55.9 \\\hline\end{array}$$
(a) Use the regression capabilities of a graphing utility to find a linear model for reaction time distance.
(b) Use the regression capabilities of a graphing utility to find a quadratic model for braking distance.
(c) Determine the polynomial giving the total stopping distance $T$.
(d) Use a graphing utility to graph the functions $R, B$, and $T$ in the same viewing window.
(e) Find the derivative of $T$ and the rates of change of the total stopping distance for $v=40, v=80$, and $v=100$
(f) Use the results of this exercise to draw conclusions about the total stopping distance as speed increases.
Differentiation
Basic Differentiation Rules and Rates of…