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Question

# 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:-3 m/s−12 m/s42 m/s−9 m/s

A
9 m/s
B
12 m/s
C
42 m/s
D
3 m/s
Solution
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#### S=t3−62+3t+4v=dsdtSo =3t2−12+3=vagain differentiate ,a=dvdt6t−12=0t=2sv=3(2)2−12×2+3=−9ms−1Hence (D) is correct answer.

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