A 325-kg boat is sailing 15.0° north of east at a speed of 2.00 m/s. Thirty seconds later, it is sailing 35.0° north of east at a speed of 4.00 m/s. During this time, three forces act on the boat: a 31.0N force directed 15.0° north of east, a 23.0N force directed 15.0° south of west and F Find the force F Answer : [18.4N, 68.0° north of east] Fr = F1 + F2 + F3 (due to 3 forces acting on it) F3 = Fr - F1 - F2 (as question wanted to look for the 3rd force) F1 = 31cos15° = 30.0N (to the right) F2 = -(23cos15) = -22.2N (to the right. Or +ve to the left) a = (v - u)/t Do I need to search for the vertical, y and the horizontal, x component of the velocity to turn it into resultant Force? Through square root of (Fy^2 +Fx^2)? I do not understand the first 2 sentences of the question. So in the end they will turn into the resultant force acted on the boat? I am confused. It is a relatively new topic in mechanics to me. Im sorry I could not even make a single line of solution.. >< i have no idea where to start calculating actually. Though I did forces of equilibrium before, this doesn't seems like anything from it.