The velocity displacement graph of a particle moving along a straight line is shown

The most suitable acceleration-displacement graph will be

Solution:

The equation for the given v-x graph is  $\mathrm{v}=-\frac{{\mathrm{v}}_0}{{\mathrm{x}}_0}\mathrm{x}+{\mathrm{v}}_0$
Differentiating the above equation w.r.t. x, we get  $\frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{{\mathrm{v}}_0}{{\mathrm{x}}_0}$
Multiplying the above equation in both sides by v, we get
 $\mathrm{v}\times \frac{\mathrm{dv}}{\mathrm{dx}}=-\frac{{\mathrm{v}}_0}{{\mathrm{x}}_0}\times \mathrm{v}=-\frac{{\mathrm{v}}_0}{{\mathrm{x}}_0}\left[-\frac{{\mathrm{v}}_0}{{\mathrm{x}}_0}\times +{\mathrm{v}}_0\right]\mathrm{From}\left(1\right)$$\mathrm{a}=\frac{{\mathrm{v}}_0^2}{{\mathrm{x}}_0^2}\times -\frac{{\mathrm{v}}_0^2}{{\mathrm{x}}_0^2}...\left(2\right)\hspace{0.17em}\hspace{0.17em}\left[\because \mathrm{a}=\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dx}}\right]$
On comparing the equation (ii) with equation of a straight line y= mx+c
We get $\mathrm{m}=\frac{{\mathrm{v}}_0^2}{{\mathrm{x}}_0^2}=+\mathrm{ve},\mathrm{i}.\mathrm{e}.\mathrm{tan\theta }=0,\mathrm{i}.\mathrm{e}.,\mathrm{\theta }$ is acute . Also $\mathrm{c}=-\frac{{\mathrm{v}}_0^2}{{\mathrm{c}}_0^2},\mathrm{i}.\mathrm{e}$ the intercept is negative. The above conditions are satisfied in graph (A)