Velocity of particle varies with position as shown in the below graph.

Find the correct variation of acceleration with position.

Based on the provided graph of velocity ($v$) versus position ($x$) in the question, the velocity decreases linearly with position. We can write the relation for velocity as:
$v=-mx+v_0$
where $m$ is the magnitude of the slope of the line ($m>0$) and $v_0$ is the positive y-intercept representing the initial velocity at $x=0$.

To find the variation of acceleration ($a$) with position ($x$), we use the relation:
$a=v\frac{dv}{dx}$

First, we differentiate the velocity equation with respect to position $x$:
$\frac{dv}{dx}=-m$

Now, substituting $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a=(-mx+v_0)(-m)$
$a=m^{2}x-m{v}_0$

Let $a_0=m{v}_0$. Since both $m$ and $v_0$ are positive constants, $a_0$ is a positive constant. Thus, the equation simplifies to:
$a=m^{2}x-a_0$

This equation represents a straight line in the $a-x$ plane with:
1. A positive slope equal to $m^2$
2. A negative vertical intercept equal to $-a_0$ at $x=0$

Therefore, the acceleration starts at a negative value $-a_0$ on the vertical axis and increases linearly, crossing the position axis into positive values. This corresponds to the graph shown in Option 3.