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Q.

The acceleration of a particle is increasing linearly with time ‘t’ as bt. The particle starts from the origin with an initial velocity $v_{0} .$ The distance travelled by the particle in time ‘t’ will be

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a

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

b

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

c

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

d

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

answer is C.

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Detailed Solution

$a = \frac{d v}{d t} = b t$
$d v = b t d t$
$\int_{v_{0}}^{v} d v = \int_{0}^{t} b t d t$
$v - v_{0} = \frac{b t^{2}}{2} \Rightarrow v = v_{0} + \frac{b t^{2}}{2} \Rightarrow \frac{d x}{d t} = v_{0} + \frac{b t^{2}}{2}$
$d x = v_{0} d t + \frac{b t^{2}}{2} d t$
$\int_{0}^{x} d x = \int_{0}^{t} v_{0} d t + \frac{b}{2}$
$x = v_{0} t + \frac{b t^{3}}{6}$

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