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

question 283

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. \(79.8 \mathrm { ft } / \mathrm { sec }\)