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Question

The accleration of a particle is given by a=(3t2+2t+2) ms2 where t is time. If the particle starts out with an initial velocity v=2 m/s at t=0, then the velocity of the particle at the end of 2 s is

A
12 m/s
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B
14 m/s
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C
16 m/s
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D
18 m/s
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Solution

The correct option is D 18 m/s
As we know that
Acceleration, a=dvdt
dv=a dt
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
(v2)=(160)
v=18 m/s

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