Consider a point $P(x, y)$ at any time $t$ on graph.
As given that initial velocity $=v_0$
Let the distance travelled in time $t=x_0$ and
Let $\angle A B O$ is $\theta$ then
$$
\tan \theta=\frac{A Q}{Q P}=\left[\frac{v_0-v}{x}\right]=\left[\frac{v_0}{x_0}\right]
$$
where, $v$ is velocity and $x$ is displacement at any instnat of time $t$.


From Eq. (i),
$$
v_0-v=\frac{v_0}{x_0} x
$$
So, $v=\frac{-v_0}{x_0} x+v_0$ ........ (i)
or, $v=v_0\left[1-\frac{x}{x_0}\right]$
It is the relation betwen $v$ and $x$.
As velocity decreases from $v_0$ to zero so acceleration is negative
$$
a=\frac{d v}{d t}=\frac{-v_0}{x_0} \frac{d x}{d t}
$$

By putting the value of $(v)$,
$$
a=\frac{-v_0}{x_0}(v)
$$
From eq, (i)
$$
=\frac{-v_0}{x_0}\left(\frac{-v_0}{x_0} x+v_0\right) \quad \therefore a=\frac{v_0^2}{x_0^2} x-\frac{v_0^2}{x_0}
$$