The acceleration of a particle increasing linearly with time t is bt. The particle starts from the origin with an initial velocity $υ_{0} .$ The distance travelled by the particle in time t will be

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$v_{0} t + \frac{1}{3} b t^{3}$

$v_{0} t + \frac{1}{3} b t^{2}$

$v_{0} t + \frac{1}{6} b t^{3}$

$v_{0} t + \frac{1}{6} b t^{2}$

answer is A.

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Detailed Solution

Given, $\frac{d v}{d t} = b t o r d v = b t d t$

$\Rightarrow \int_{v_{0}}^{v} d v = \int_{0}^{t} b t d t$

$o r v - v_{0} = \frac{b t^{2}}{2}$

$o r v = v_{0} + \frac{b t^{2}}{2}$

$o r d x = v_{0} d t + \frac{b t^{2}}{2} d t$

$\int_{0}^{x} d x = \int_{0}^{t} v_{0} d t + \frac{b}{2} \int_{0}^{t} t^{2} d t$

$o r x = v_{0} t + \frac{1}{2} \frac{b t^{3}}{3} = v_{0} t + \frac{b t^{3}}{6}$

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