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

# 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 is

1. A

t=0 only

2. B

t=8 only

3. C

t = 0, 8

4. D

none of these

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### Solution:

We have ,

$s=6{t}^{2}-\frac{{t}^{3}}{2}⇒\frac{ds}{dt}=12t-\frac{3{t}^{2}}{2}$ and $\frac{{d}^{2}s}{d{t}^{2}}=12-3t$

When the particle is momentarily at rest, we have

$\frac{ds}{dt}=0$ and $\frac{{d}^{2}s}{d{t}^{2}}\ne 0$

Now,

$\frac{ds}{dt}=0⇒t=0,t=8$

Clearly, $\frac{{d}^{2}s}{d{t}^{2}}\ne 0$ for $t=0,8$ .

Thus, the particle is momentarily at rest at

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