If initial velocity of a particle is $u$ (at $t = 0$ ) and the acceleration is $f = a t$ , then

v = u + a t^{2}

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v = u + \frac{a t^{2}}{2}

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v = u + a t

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v = u

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Solution

The correct option is C $v = u + \frac{a t^{2}}{2}$
Acceleration = $a t$
Increase in velocity
$= A r e a o f Δ O A B$
$= \frac{1}{2} \times a t \times t$
$= \frac{1}{2} a t^{2}$
$∴ v = u + \frac{1}{2} a t^{2}$

Also, by integration method:
Given acceleration, $f = a t$
Acceleration $=$ rate of change of velocity
$f = \frac{d v}{d t} = a t$
Integrating with respect to time , we get
$\int_{u}^{v} d v = \int_{0}^{t} a t d t$
$v - u = \frac{a t^{2}}{2}$
$v = u + \frac{a t^{2}}{2}$

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