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Velocity

The position ...

Question

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

\frac{a}{b}

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\frac{2 a}{3 b}

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\frac{a}{3 b}

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Zero

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Solution

The correct option is C $\frac{a}{3 b}$
$x = a t^{2} - b t^{3}$
$\Rightarrow$ Velocity $= \frac{d x}{d t} = 2 a t - 3 b t^{2}$
Acceleration $= \frac{d^{2} x}{d t^{2}} = \frac{d}{d t} (2 a t - 3 b^{2}) = 2 a - 6 b t$
or $a = 2 a - 6 b t$
Making $a = 0$
$\Rightarrow 2 a = 6 b t$
$⇒t=dfrac2a6b=a3b$ .

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