If a particle moves according to the law- s-6t2 - t3-2 - then the time at which it is momentarily at rest

The correct option is C t=0,8 s=6t^2 t^32 v=dsdt=12t 3t^22 Since, v=0 ⇒ 3t(8 t)=0 ⇒ t=0,8 ...

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Standard XII

Physics

Differentiation

If a particle...

Question

If a particle moves according to the law, $s = 6 t^{2} - \frac{t^{3}}{2}$ , then the time at which it is momentarily at rest

t = 0

only

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t = 8

only

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t = 0, 8

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none of these

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Solution

The correct option is C $t = 0, 8$
$s = 6 t^{2} - \frac{t^{3}}{2}$
$v = \frac{d s}{d t} = 12 t - \frac{3 t^{2}}{2}$
Since, $v = 0$
$\Rightarrow 3 t (8 - t) = 0$
$\Rightarrow t = 0, 8$

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If a particle moves according to the law, s=6t2−t32 , then the time at which it is momentarily at rest

The correct option is C t=0,8s=6t2−t32v=dsdt=12t−3t22Since, v=0⇒3t(8−t)=0⇒t=0,8

If a particle moves according to the law, $s = 6 t^{2} - \frac{t^{3}}{2}$ , then the time at which it is momentarily at rest

The correct option is C $t = 0, 8$
$s = 6 t^{2} - \frac{t^{3}}{2}$
$v = \frac{d s}{d t} = 12 t - \frac{3 t^{2}}{2}$
Since, $v = 0$
$\Rightarrow 3 t (8 - t) = 0$
$\Rightarrow t = 0, 8$