$\text { Suppose that an automobile's velocity starting from rest is } v ( t ) = \frac { 240 t } { 5 t + 13 } \text { where } v \text { is }$ measured in feet per second. Find the acceleration at 9 seconds. Round your answer to one decimal place.
A) $1.9 \mathrm { ft } / \mathrm { sec } ^ { 2 }$
B) $0.9 \mathrm { ft } / \mathrm { sec } ^ { 2 }$
C) $0.6 \mathrm { ft } / \mathrm { sec } ^ { 2 }$
D) $0.2 \mathrm { ft } / \mathrm { sec } ^ { 2 }$
E) $8.3 \mathrm { ft } / \mathrm { sec } ^ { 2 }$

Question

$\text { Suppose that an automobile's velocity starting from rest is } v ( t ) = \frac { 240 t } { 5 t + 13 } \text { where } v \text { is }$ measured in feet per second. Find the acceleration at 9 seconds. Round your answer to one decimal place. 
A) $1.9 \mathrm { ft } / \mathrm { sec } ^ { 2 }$
B) $0.9 \mathrm { ft } / \mathrm { sec } ^ { 2 }$
C) $0.6 \mathrm { ft } / \mathrm { sec } ^ { 2 }$
D) $0.2 \mathrm { ft } / \mathrm { sec } ^ { 2 }$
E) $8.3 \mathrm { ft } / \mathrm { sec } ^ { 2 }$

Quizplus · Accepted Answer

Acceleration is the derivative of velocity with respect to time. The derivative of $v(t) = \frac{240t}{5t + 13}$ is $a(t) = v'(t) = \frac{240(5t + 13) - 240t(5)}{(5t + 13)^2} = \frac{3120}{(5t + 13)^2}$. Substituting $t = 9$ gives $a(9) \approx 0.9 \, \text{ft/sec}^2$.