The slope of kinetic energy and displacement curve for a particle in motion will be [BCECE Mains 2012] 

Kinetic energy of a particle in motion,  $E=\frac{1}{2}m{v}^2$

Differentiating above equation w.r.t. x, we get

$\frac{dE}{dx}=\frac{1}{2}m\times 2v\frac{dv}{dx}=mv\times \frac{dv}{dt}\times \frac{dt}{dx}=mv\times \frac{a}{v}=ma$

So, the slope of kinetic energy-displacement curve is directly proportional to the acceleration of the particle.