JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Kinematics

The velocity-displacement graph rises from $10\mathrm{m}{\mathrm{s}}^{-1}$ at the origin to $50\mathrm{m}{\mathrm{s}}^{-1}$ at $x=200$ m and then stays flat, so

Acceleration along a straight path follows from $a=v\frac{dv}{dx}$. On the sloping stretch $\frac{dv}{dx}=\frac{1}{5}$, so

which is a straight line of positive slope starting at $2\mathrm{m}{\mathrm{s}}^{-2}$. Beyond $x=200$ m the speed no longer changes, $\frac{dv}{dx}=0$, and the acceleration falls to zero.
The acceleration-displacement graph is a straight line rising from $2\mathrm{m}{\mathrm{s}}^{-2}$ over the first 200 m, dropping abruptly to zero there and staying at zero out to 400 m.