An object is moving so that its velocity after t minutes is $v ( t ) = 4 + 6 t + 9 t ^ { 2 }$ meters per minute. How far does the object travel from the end of minute 4 to the end of minute 5?
A) 214 meters
B) 690 meters
C) 210 meters
D) 156 meters

Question

Quizplus · Accepted Answer

To find the distance traveled from the end of minute 4 to the end of minute 5, we need to integrate the velocity function $v(t) = 4 + 6t + 9t^2$ from $t = 4$ to $t = 5$. The integral of $v(t)$ gives the displacement function $s(t)$, which is $s(t) = \int v(t) dt = \int (4 + 6t + 9t^2) dt = 4t + 3t^2 + 3t^3 + C$. To find the distance traveled, we evaluate $s(5) - s(4)$.$s(5) = 4(5) + 3(5)^2 + 3(5)^3 = 20 + 75 + 375 = 470$$s(4) = 4(4) + 3(4)^2 + 3(4)^3 = 16 + 48 + 192 = 256$Therefore, the distance traveled from the end of minute 4 to the end of minute 5 is $470 - 256 = 214$ meters.

An Object Is Moving So That Its Velocity After T v(t)=4+6t+9t2v ( t ) = 4 + 6 t + 9 t ^ { 2 }v(t)=4+6t+9t2

An Object Is Moving So That Its Velocity After T $v ( t ) = 4 + 6 t + 9 t ^ { 2 }$