The displacement $(x)$ of a particle starting from rest is given by $x = 6 t^{2} - t^{3}$ . The time at which the particle will attain zero velocity again is

2

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4

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6

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8

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Solution

The correct option is B $4$

Velocity, $v = \frac{d x}{d t} = 6 (2 t) - 3 t^{2}$
$= 12 t - 3 t^{2}$

$= 3 t (4 - t)$

Now, if $v = 0 \Rightarrow t = 0 a n d t = 4$

Suggest Corrections

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