Word for word: A50,000kg locomotive is traveling at 10 m/s when its engine and breaks fail. How far will the locomotive roll before it comes to a stop? Rolling friction for rubber on concrete is 0.02 (by the book) a= fnet/m = (umg)/m so mass cancels out leaving a = ug = .02(-9.8) = -.196 I use Vf^2 = Vi^2 + 2ad (I'll just say d = distance in meters) So it works out to -100 = (-0.392)d d = 255 m but wait... the book says the answer's 2550! I would figure that means the rolling friction is .002 but that's steel on steel (dry) What am I doing wrong???