Answer
curve-C is $s(t)$, curve-B is $v(t)$, and curve-A is $a(t)$,
Work Step by Step
Step 1. Establish the relations: base curve $s(t)$, curve $v(t)=s'(t)$, curve $a(t)=v'(t)=s''(t)$
Step 2. Using the derivative property that a horizontal tangent line gives a value of zero to the derivative of that curve, we can identify that the zero in curve-A is a derivative of the minimum of curve-B, thus $A=B'$
Step 3. The zeros of curve-B at endpoints comes from the derivatives of curve-C because the slopes of the tangent lines at the endpoints of curve-C are zeros, thus $B=C'$
Step 4. We conclude that curve-C is $s(t)$, curve-B is $v(t)$, and curve-A is $a(t)$,
