The given graph shows the variation of velocity (v) with position...

(A)
Given line have positive intercept but negative slope.
So its equation can be written as,
(v=-m x+v_{0} \ldots(1))
where (m=\tan $\theta=\frac{v_{0}}{x_{0}}$
By differentiating with respect to time we get, ($\frac{d v}{d t}=-$m $\frac{d x}{d t}=-$m v)
Now substituting the value of ' $v$ ' from equation (1) we get
($\frac{d v}{d x}=-$m\left[-m x+v_{0}\right]=m^{2} x-m v_{0}) $a=m^{2} x-m v_{0}$
i.e., the graph between a and $x$ should have positive slope but negative intercept on acceleration axis.
Hence option (A) is correct.