AP Calculus ABeasymcq1 pt

A particle moving along a line has acceleration a(t)=6ta(t)=6t and initial velocity v(0)=2v(0)=2. What is v(3)v(3)?

A.2727
B.2929
C.5656
D.55

Explanation

Core Concept

By the Fundamental Theorem, v(3)=v(0)+036tdt=2+[3t2]03=2+27=29v(3)=v(0)+\int_0^3 6t\,dt=2+[3t^2]_0^3=2+27=29.

Correct Answer

B2929

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