Physics - Kinematics Question with Solution | TestHub

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

v²= 4 + 4s

Now, comparing it with  v²= u²+ 2as

we get,    u = 2 m/s and a = 2 m/s²

$\therefore$Displacement at t = 2s   is

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

or    s = 8 m