The question reads as such: A car of mass 1500kg traveling at 24 m/s is at the foot of a hill that rises 120 m in 2.0 km. At the top of the hill the speed of the car is 10 m/s. Find the average power delivered by the car's engine, neglecting any frictional losses. The book says the answer is 18 kW. My work gets me the answer 11980.8 W, my friend got 15 kW. Here's what I did: The actual hill is 2003.6 m long (Pythagorean). The angle of incline is 3.44 deg (sin^-1(120/2003.6)). Acceleration due to gravity is -9.8 sin(3.44), or -.588 m/s^2. The car decelerates from 24 to 10 m/s, so at = -14. d = 1/2 at^2 + vi*t Sub in at, d = 1/2 (-14)t + vi*t This is the step that might be wrong, I suppose, because this idea just came to me out of the blue. It SHOULD work, since a*t is a*t regardless, but meh, I've been wrong before. 2003.6 = .5(-14)t + (24)t t = 117.9 at = -14 a = -14/117.9 = -.118 a(gravity) + a(engine) = a(total) -.588 + a(engine) = -.118 a(engine) = .47 m/s^2 f(engine) = ma f(engine) = 1500 * .47 = 705 N p(engine) = (f*d)/t p(engine) = (705 * 2003.6)/117.9 = 11980.8 W Any insight into this problem?