#### Answer

$(C)$ represent position $s$, $(B)$ represent velocity $v$ and $(A)$ represent acceleration $a$.

#### Work Step by Step

To recognize the curve $(a)$ and its derivative curve $(a')$, what I would do first is to look for corresponding points: change-of-sign points in $(a')$ should correspond to change-of-direction points in $(a)$. The reason is that
- As $a'\gt0$, the graph of $a$ moves upward.
- As $a'\lt0$, the graph of $a$ moves downward.
So a change in the sign of $a'$ will signify a change in the direction of $(a)$ at that exact same point.
The blue curve $(A)$ and the yellow curve $(B)$ represent such a pair. At the point $(A)$ changes sign from negative to positive, $(B)$ changes from moving downward to upward. So $(A)$ is the derivative of $(B)$.
But still that leaves the relationship between the red curve $(C)$ and the other two in question. A clue we can see in $(C)$ is that it is continuously moving downward. The derivative of $(C)$, as a result, must be negative all the time, which we can relate to the curve $(B)$.
Another reason to believe that $(B)$ is the derivative $(C)$ is that for the first half, $(C)$ decreases more and more steeply, corresponding to the downward trend in $(B)$. For the second half, $(C)$ decreases less and less steeply, corresponding to the upward trend in $(B)$.
Therefore, $(C)$ represent position $s$, $(B)$ represent velocity $v$ and $(A)$ represent acceleration $a$.