I would draw you a force diagram to help with the explanation, but I don't know how to put a picture up. Therefore I'll just try my best without.
The problem comes in two phases. The first is when the jets are operating. Here the forward horizontal force is 200N, and there is a frictional force Fr acting in antiparallel (the opposite direction).
Forces are balanced in the vertical sense, since the skier neither takes off from or sinks into the water. Therefore his normal reaction force, R, is equal to his weight force. Since W = mg, we see that R = mg.
Next, use the formula Fr = uR, where u is the coefficient of kinetic friction, to calculate Fr = umg.
This makes the resultant horizontal force in phase 1 equal to (200 - umg)N. By Newton's Second Law of Motion, and with the slightly callous assumption that the combined mass of him and his apparatus is constant, F = ma. Therefore a = F/m = ((200 - umg)/m) ms^-1.
Now we can use the equations of motion for constant acceleration to calculate both the skier's velocity and displacement at t = 94s.
S = ?
U = 0ms^-1
V = ?
a = ((200 - umg)/m) ms^-2
t = 94s
Using S = Ut + 0.5at^2:
S = (0)(94) + (0.5)((200-umg)/m)(94^2) =
