Potential energy of a particle along x-axis varies as shown in the figure. If the particle is released from rest at a point on x-axis.

$\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}$if slope is +Ve body will move in the negative X direction. Also, since the slope is constant acceleration is constant. Thus $\frac{\mathrm{dv}}{\mathrm{dt}}$= constant and as a result $\frac{\mathrm{dv}}{\mathrm{dx}}\frac{\mathrm{dx}}{\mathrm{dt}}$ = constant.$\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{V}$ = constant. On integrating, $\frac{{\mathrm{V}}^2}{2}$= X + C. Thus V² is $\propto$X.