Solve the problem.
-The driver of a car traveling at $36 \mathrm { ft } / \mathrm { sec }$ suddenly applies the brakes. The position of the car is $s = 36 \mathrm { t } - 3 \mathrm { t } ^ { 2 }$, $t$ seconds after the driver applies the brakes. How far does the car go before coming to a stop?
A) 432 ft
B) 108 ft
C) 216 ft
D) 6 ft

Question

Quizplus · Accepted Answer

The car comes to a stop when its velocity is 0. The velocity is the derivative of the position function $s(t) = 36t - 3t^2$, which is $s'(t) = 36 - 6t$. Setting this equal to 0 gives $36 - 6t = 0$, solving for $t$ gives $t = 6$ seconds. Substituting $t = 6$ into the position function gives $s(6) = 36(6) - 3(6)^2 = 216 - 108 = 108$ ft.

Solve the Problem 36ft/sec36 \mathrm { ft } / \mathrm { sec }36ft/sec

Solve the Problem $36 \mathrm { ft } / \mathrm { sec }$