The velocity of the center of mass of a system of particles changes as v = 4.5 t + 2.4 t² + 1.1 t³, where v is in meters per second.If the system starts from rest at t = 0, what is its position at t = 3.0 s? A)7.1 m B)21 m C)40 m D)49 m E)64 m

Question

Quizplus · Accepted Answer

To find the position, integrate the velocity function with respect to time. The integral of v = 4.5t + 2.4t^2 + 1.1t^3 is s(t) = (4.5/2)t^2 + (2.4/3)t^3 + (1.1/4)t^4 + C. Since the system starts from rest, C = 0. Evaluating at t = 3.0 s gives s(3) = (4.5/2)(3)^2 + (2.4/3)(3)^3 + (1.1/4)(3)^4 = 64 m.

The Velocity of the Center of Mass of a System