The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with an initial velocity v₀. The distance travelled by the particle in time t will be

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

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

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

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

answer is C.

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Detailed Solution

$a = bt \Rightarrow \frac{dv}{dt} = bt$
Integrating both sides we get $\int_{v_{0}}^{v} dv = \int_{0}^{t} btdt$
$\begin{array}{l} v = v_{0} = \frac{1}{2} {bt}^{2}; v = v_{0} + \frac{1}{2} {bt}^{2} \\ \frac{ds}{dt} = v_{0} + \frac{{bt}^{2}}{2}; \int_{0}^{s} ds = \int_{0}^{t} (v_{0} + \frac{{bt}^{2}}{2}) dt s = v_{0} t + \frac{{bt}^{3}}{6} \end{array}$