Solve the problem.
-A car accelerates from 0 to $91.2 \mathrm { ft } / \mathrm { sec }$ in $8 \mathrm { sec }$. The distance $d ( t )$ (in $\mathrm { ft }$ ) that the car travels $t$ seconds after motion begins is given by $d ( t ) = 5.7 t ^ { 2 }$, where $0 \leq t \leq 8$.
a. Find the difference quotient $\frac { d ( t + h ) - d ( t ) } { h }$.
b. Use the difference quotient to determine the average rate of speed on the interval $4 \leq \mathrm { t } \leq 6$.
A) a. $5.7 ( t + h )$;
b. $34.2 \mathrm { ft } / \mathrm { sec }$
B) a. $11.4 t + 5.7 h$;
b. $57 \mathrm { ft } / \mathrm { sec }$
C) a. $5.7 ( t + h )$;
b. $79.8 \mathrm { ft } / \mathrm { sec }$
D) a. $11.4 t + 5.7 h$;
b. $79.8 \mathrm { ft } / \mathrm { sec }$

Question

Quizplus · Accepted Answer

a. The difference quotient $\frac{d(t+h)-d(t)}{h}$ simplifies to $\frac{5.7(t+h)^2-5.7t^2}{h}$, which further simplifies to $11.4t + 5.7h$. b. For the interval $4 \leq t \leq 6$, the average rate of speed is the change in distance over the change in time, which is $\frac{d(6)-d(4)}{6-4}$, resulting in $57 \mathrm { ft } / \mathrm { sec }$.

Solve the Problem 91.2ft/sec91.2 \mathrm { ft } / \mathrm { sec }91.2ft/sec

Solve the Problem $91.2 \mathrm { ft } / \mathrm { sec }$