Solve the problem.
-For a motorcycle traveling at speed $v$ (in $\mathrm{mph}$ ) when the brakes are applied, the distance $d$ (in feet) required to stop the motorcycle may be approximated by the formula $d=0.05 v^{2}+v$. Find the instantaneous rate of change of distance with respect to velocity when the speed is $44 \mathrm{mph}$.
A) $5.4 \mathrm{mph}$
B) $10.8 \mathrm{mph}$
C) $4.4 \mathrm{mph}$
D) $45 \mathrm{mph}$

Question

Quizplus · Accepted Answer

The instantaneous rate of change of distance with respect to velocity is the derivative of the distance function with respect to velocity. The derivative of $d=0.05 v^{2}+v$ with respect to $v$ is $d'=0.1 v+1$. When the speed is $44 \mathrm{mph}$, the instantaneous rate of change of distance with respect to velocity is $d'(44)=0.1 (44)+1=5.4 \mathrm{mph}$.

Solve the Problem vvv (In mph\mathrm{mph}mph ) When the Brakes Are Applied, the Distance

Solve the Problem $v$ (In $\mathrm{mph}$ ) When the Brakes Are Applied, the Distance