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Acceleration

The position ...

Question

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 equal to zero, at time

\frac{2 a}{3 b}

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\frac{a}{b}

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\frac{a}{3 b}

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\frac{2 a}{b}

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Solution

The correct option is C $\frac{a}{3 b}$
$x = a t^{2} - b t^{3} \frac{d x}{d t} = v = 2 a t - 3 b t^{2} \frac{d^{2} x}{d t^{2}} = a = 2 a - 6 b t = 0 t = \frac{a}{3 b}$

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