A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6 t^{2} + 3 t + 4$ metres. The velocity when the acceleration is zero is:-

3 m / s

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- 12 m / s

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42 m / s

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- 9 m / s

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Solution

The correct option is D $- 9 m / s$
$S = t^{3} - 6^{2} + 3 t + 4$
$v = \frac{d s}{d t}$

So $= 3 t^{2} - 12 + 3 = v$
again differentiate , $a = \frac{d v}{d t}$
$6 t - 12 = 0$
$t = 2 s$
$v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$

Hence (D) is correct answer.

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A particle moves along a straight line such that its displacement(s) at any time(t) is given by $S = t^{3} - 3 t^{2} + 2$ . The velocity of the particle when its acceleration zero is?

Q. A particle is moving so that its displacement s s given as

s = t^{3} - 6 t^{2} + 3 t + 4

meter. Its velocity at the instant when its acceleration is zero will be -

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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:-

The correct option is D −9 m/sS=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.

A particle moves along a straight line such that its displacement at any time $t$ is given by $s = t^{3} - 6 t^{2} + 3 t + 4$ metres. The velocity when the acceleration is zero is:-

The correct option is D $- 9 m / s$
$S = t^{3} - 6^{2} + 3 t + 4$
$v = \frac{d s}{d t}$

So $= 3 t^{2} - 12 + 3 = v$
again differentiate , $a = \frac{d v}{d t}$
$6 t - 12 = 0$
$t = 2 s$
$v = 3 (2)^{2} - 12 \times 2 + 3 = - 9 m s^{- 1}$

Hence (D) is correct answer.