Chapter 1.8, Problem 123PE

A car traveling 60 mph (88 ft/sec) undergoes a constant deceleration until it comes to rest approximately 9.09 sec later. The distance $d\left(t\right)$ (in ft) that the car travels t seconds after the brakes are applied is given by $d\left(t\right)=-4.84t^2+88t,\text{ where }0\le t\le \mathrm{9.09.}$ (See Example 5)

a. Find the difference quotient $\frac{d\left(t+h\right)-d\left(t\right)}{h}.$ Use the difference quotient to determine the average rate of speed on the following intervals for t.

$\begin{array}{l}\text{b}\text{. }\left[0,2\right]\\ \text{c}\text{. }\left[2,4\right]\\ \text{d}\text{. }\left[4,6\right]\\ \text{e}\text{. }\left[6,8\right]\end{array}$