Question

The velocity-position graph of a particle is shown in figure. Obtain the relation between acceleration and displacement and plot it.

Answer 1

As it is a straight line graph showing deceleration

We can find

$\tan \theta =\frac{v_0-v}{x}$

$\frac{v_0}{x_0}=\frac{v_0-v}{x}$

So the velocity is $-\frac{v_0}{x_0}x+v_0$

The acceleration will be

$a=\frac{dv}{dt}$

$a=\frac{-v_0}{x_0}v$

$=\frac{-v_0}{x_0}\left(\frac{-v_0}{x_0}x+v_0\right)$

Thus the relation between the acceleration and displacement will be a straight line having intercept$\frac{-v_0^2}{x_0}$