JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Kinematics

The velocity falls linearly from $v_0$ at $x=0$ to zero at $x=x_0$, so $v=v_0\left(1-\frac{x}{x_0}\right)$. Acceleration as a function of displacement follows from $a=v\frac{dv}{dx}$:

This is linear in $x$, equal to $-\frac{v_0^2}{x_0}$ at $x=0$ and rising to zero at $x=x_0$. The acceleration-displacement graph is therefore a straight line of positive slope that starts below the origin on the acceleration axis and meets the displacement axis at $x_0$.