The question refers to a velocity–displacement graph, and asks for the corresponding acceleration–displacement graph.

However, the actual figure and the option graphs are not visible in the provided prompt. So the exact graph cannot be identified from the given information alone.

Still, the relation needed to convert a $v$–$x$ graph into an $a$–$x$ graph is:

1. Key relation

We use $a = \frac{dv}{dt}$ and since $\frac{dv}{dt} = \frac{dv}{dx}\cdot \frac{dx}{dt} = v\frac{dv}{dx},$ we get $\boxed{a(x)=v(x)\,\frac{dv}{dx}}.$

So, from a given velocity–displacement graph:

1. Read $v$ at each displacement $x$.
2. Find the slope $\dfrac{dv}{dx}$ at that point.
3. Multiply them: $a=v\frac{dv}{dx}.$

2. How the graph would be obtained

• If the $v$–$x$ graph is a straight line with positive slope, then $v = mx + c$ so $a = v\frac{dv}{dx} = (mx+c)m = m^2x+mc,$ which is a straight line in $a$–$x$.

• If the $v$–$x$ graph is horizontal, then $\frac{dv}{dx}=0 \implies a=0.$

• If the $v$–$x$ graph decreases linearly, then $\dfrac{dv}{dx}<0$, so the sign of acceleration depends on the sign of $v$.

3. Limitation here

Because the actual figure and option graphs are missing, I cannot independently derive which of A, B, C, or D matches.

The stored correct answer is A, but without the graph, this cannot be verified from first principles.

So the best I can do is state the governing relation: $\boxed{a=v\frac{dv}{dx}}$ and note that the exact option depends on the unseen figure.