AP Calculus ABmediummcq1 pt

A particle moves along a line with velocity v(t)=3t2+2v(t)=3t^2+2 meters per second. What is the displacement of the particle over the interval [1,3][1,3]?

A.3030 meters
B.2626 meters
C.2828 meters
D.3333 meters

Explanation

Core Concept

Displacement is 13(3t2+2)dt=[t3+2t]13=(27+6)(1+2)=30\int_1^3(3t^2+2)\,dt=[t^3+2t]_1^3=(27+6)-(1+2)=30.

Correct Answer

A3030 meters

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