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

Multiple Choice

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. $79.8 \mathrm { ft } / \mathrm { sec }$