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The position x of a particle varies with time t as $x = a t^{2} - b t^{3} .$ The acceleration of the particle will be zero at time equal to

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$\frac{a}{b}$

$\frac{2 a}{3 b}$

$\frac{a}{3 b}$

Zero

answer is C.

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Detailed Solution

$v = \frac{d x}{d t} = 2 a t - 3 b t^{2}$

$a c c = \frac{d v}{d t} = 2 a - 6 b t$

$a c c = \frac{a}{3 b}$

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