There is a single cable beginning at the ceiling and passing through the double

Question

There is a single cable beginning at the ceiling and passing through the double pulley C up to D down to A back up to C and down to B and back C as shown in the figure. Starting from t=0, for the next two seconds, the distance that A is from the ceiling, in inches, is 6t+150 if t is measured in seconds. Similarly, the distance that B is from the ceiling, in inches, is -2t^5+200. Find the velocity of the double pulley C after one second.

Im really lost on this problem. Can you please list the formulas being used and how you got them. I dont understand this problem at all. Thank You!! I will give you a lifesaver if you can do this.

Explanation / Answer

if we use yA, yB, yC to represent the y coordinates of A, B, C, respectively, where +y is down and the origin is at the ceiling, then the total length of the cable is 2yC + yA + (yA - yC) + 2(yB - yC) = L = constant so yC = 2yA + 2yB - L = 2(6t + 150) + 2(-2t5 + 100) - L vC = dyC/dt = 12 - 20t4 at t = 1 s, vC = -8 in/s

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