A car is traveling at 40.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s/s. If the coefficient of friction between road and tires on a rainy day is 0.12, what is the minimum distance in which the car will stop? (1 mi = 1.609 km) So given: vi(initial velocity): 40.7 mi/h -> 65486.3m/3600s vf(final velocity): 0 m/s g(gravity): -9.8 Ff(friction): -0.12 m(mass): ? Fnet(net force): ? x(distance): ? Equations: F = ma Fnet = ma - Ff Ff = uFn Fn = mg vf(^2) = vi(^2) + 2ax Answer in back of text book: 140.689 m I am unsure how to set this problem up. All my attempts have had no success. Please, help would be appreciated.