The question: The driver of a 1.2X10^3 kg car travelling 45 km/h [W] on a slippery road applies the brakes, skidding to a stop in 35 m. Determine the coefficient of friction between the road and the car tires. Equations: - Ff = uFn - d = 0.5 (Vi + Vf) * time - Fn = mg - F = ma My attempt: To find normal force: Fn = (1.2 * 10^3) (9.81) = 11772 N To find acceleration: d = 0.5 (Vi + Vf) * time 35 = 0.5 (45 + 0) * time 35 = 0.5 (45) * time 35 = 22.5 * time 35/22.5 = time time = 1.6 seconds acceleration = (Vf - Vi) / time = (0 - 45) / 1.6 = -28.12 m/s^2 To find force using Newtons' law: F= (1.2*10^3)(.28.12) = -33744 N u (coefficient of friction) = -33744 / 11772 = -2.87 Is the answer correct? Basically we have to find the coefficient of friction, but I am confused as to how to find it out. The main problem is that acceleration is negative, can anyone please help. This is due tomorrow for me. Thanks.