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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 at time $t$ of the particle will be equal to zero. What is the value of $t$ ?

\frac{4 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}{3 B}

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Solution

The correct option is C $\frac{A}{3 B}$
Given, $x = A t^{2} - B t^{3}$
$∴$ velocity $= \frac{d x}{d t} = 2 A t - 3 B t^{2}$
and acceleration $= \frac{d}{d t} (\frac{d x}{d t}) = 2 A - 6 B t$
For acceleration to be zero = $2 A - 6 B t = 0$
$∴ t = \frac{2 A}{6 B} = \frac{A}{3 B}$

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