The accleration of a particle is given by a=(3t2+2t+2)ms−2 where t is time. If the particle starts out with an initial velocity v=2m/s at t=0, then the velocity of the particle at the end of 2s is
A
12m/s
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B
14m/s
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C
16m/s
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D
18m/s
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Solution
The correct option is D18m/s As we know that Acceleration, a=dvdt ⇒dv=adt Let the velocity of the particle at end of 2 sec is v, ∴∫v2dv=∫20(3t2+2t+2)dt ⇒[v]v2=[t3+t2+2t]20 ⇒(v−2)=(16−0) ⇒v=18m/s