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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Solution

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

$v = \frac{d s}{d t} = 3 t^{2} - 12 t + 3$

$a = \frac{d v}{d t} = 6 t - 12$

So, acceleration will be zero at
$6 t - 12 = 0$
$t = 2 s e c$
Then velocity at t = 3
$v = 3 (2)^{2} - 12 (2) + 3$
= 12-24+3
$v = - 9 m / s$

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A particle is moving so that its displacement s s given as s=t3−6t2+3t+4 meter. Its velocity at the instant when its acceleration is zero will be -

s=t3−6t2+3t=4v=dsdt=3t2−12t+3a=dvdt=6t−12So, acceleration will be zero at6t−12=0t=2secThen velocity at t = 3v=3(2)2−12(2)+3 = 12-24+3v=−9m/s

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 -

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

$v = \frac{d s}{d t} = 3 t^{2} - 12 t + 3$

$a = \frac{d v}{d t} = 6 t - 12$

So, acceleration will be zero at
$6 t - 12 = 0$
$t = 2 s e c$
Then velocity at t = 3
$v = 3 (2)^{2} - 12 (2) + 3$
= 12-24+3
$v = - 9 m / s$