The velocity-displacement graph of a particle is as shown in the figure.

 

 

 

 

 

 

 

The acceleration – displacement graph of the same partied is represented by

The slope of the given graph,

$m=\frac{-v_0}{x_0}$

So, $v=\left(\frac{-v_0}{x_0}\right)x+v_0$ and

$a=\frac{vdv}{dx}$

$a=\left[\left(\frac{-v_0}{x_0}\right)x+v_0\right]·\frac{d}{dx}\left[\left(\frac{-v_0}{x_0}x+v_0\right)\right]$

$a=\left[\left(\frac{-v_0}{x_0}\right)x+v_0\right]·\left[\left(\frac{-v_0}{x_0}+0\right)\right]$

$=\frac{v_0^2}{x_0^2}x-\frac{v_0^2}{x_0}$

Now, comparing with the standard equation of a straight line $(y=mx+c)$

Here, $m=+ve and c=-ve$