Velocity of a particle moving in a straight line varies with its...

as $a = \dfrac{vdv}{ds} = \sqrt{4 + 4s}\dfrac{1}{2\sqrt{\left(4 + 4s\right)}} \times 4 = 2,$
which is constant so we can use equation of motion now Squaring the given equation, we get

v^{2} = 4 + 4s

Now, comparing it with v^{2} = u^{2} + 2as

we get, u = 2 m/s and a = 2 m/s^{2}

$∴$ Displacement at t = 2s is

$\text{s} = \text{ut} + \dfrac{1}{2} {at}^{2}$ or $\text{s} = \left(2\right) \left(2\right) + \dfrac{1}{2} {\left(2\right) \left(2\right)}^{2}$

or s = 8 m