Chapter 2.8, 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 $d\left(t\right)\left(\text{in ft}\right)$ that the car travels t seconds after the brakes are applied is given by $d\left(t\right)=-4.84t^2+88t$, where $0\le t\le 9.09$. (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.

b. $\left[0,2\right]$ (Hint: $t=0\text{ and }h=2$

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c. $\left[2,4\right]$ (Hint: $t=2\text{ and }h=2$

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d. $\left[4,6\right]$ (Hint: $t=4\text{ and }h=2$

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e. $\left[6,8\right]$ (Hint: $t=6\text{ and }h=2$

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