Answer
curve-C is $s(t)$, curve-A is $v(t)$, and curve-B is $a(t).$
Work Step by Step
Step 1. Establish the relations among the curves: one is the position $s(t)$, another one is the velocity $v(t)=s'(t)$, and the third one is the acceleration $a(t)=v'(t)=s''(t)$
Step 2. While there may be other ways to identify the curves, we use a fundamental property of derivatives: the value of the derivative will be zero when the tangent to the original curve is horizontal.
Step 3. We can see that curve-B has one zero in the middle of the time interval and only curve-A has a horizontal tangent at that time (minimum of the blue curve); thus curve-B is a derivative of curve-A.
Step 4. Similarly, curve-A contains two zeros (cross the t-axis) corresponding to the horizontal tangents of the red curve-C; thus curve-A is a derivative of curve-C.
Step 5. Based on the above results, we can conclude that curve-C is $s(t)$, curve-A is $v(t)$, and curve-B is $a(t).$
