The acceleration of a particle is increasing linearly with time $t$ as $b t$ . The particle starts from rest from the origin. The velocity will be proportional to

t

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t^{2}

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t^{3}

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t^{0}

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Solution

The correct option is B $t^{2}$
As we know that $\frac{d v}{d t} = a$
So using the above result,
$\int_{0}^{v} d v = \int_{0}^{t} a d t = \int_{0}^{t} (b t) d t$
$\Rightarrow v - 0 = \frac{b t^{2}}{2} - 0$
$\Rightarrow v = \frac{b t^{2}}{2}$

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The correct option is B $t^{2}$
As we know that $\frac{d v}{d t} = a$
So using the above result,
$\int_{0}^{v} d v = \int_{0}^{t} a d t = \int_{0}^{t} (b t) d t$
$\Rightarrow v - 0 = \frac{b t^{2}}{2} - 0$
$\Rightarrow v = \frac{b t^{2}}{2}$