A particle moves along a straight line such that its displacement at any time t is given by s=t3−6t2+3t+4 metres. The velocity when the acceleration is zero is:-

3m/s

−12m/s

42m/s

−9m/s

A

3m/s

B

−12m/s

C

42m/s

D

−9m/s

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Solution

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S=t3−62+3t+4

v=dsdt

So =3t2−12+3=v

again differentiate ,a=dvdt

6t−12=0

t=2s

v=3(2)2−12×2+3=−9ms−1

Hence (D) is correct answer.

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