# Total distance a car moves with two coefficients of friction?

A 1000 kg car is moving at 10 m/s, the car applies the brakes and begins to skid and leave a mark. the first 4 m of the surface has a coefficient of 0.35 and after that the coefficient of friction is 0.2. What is the total distance the car will slide when coming to rest?

Equations: F=ma and F(f)=μN

Δχ=V_{0}t + 1/2at^{2}

Vf=V_{0} + at

W=Fxd

W=1/2mV_{F}^2 – 1/2mV_{0}^2

Since the car is going to rest, I’m going to assume the final velocity is 0. Since there are no other Y axis forces, I think you can assume N=mg so it equals 9800N. I tried using the F=ma of the x components and got -(9800*.35)+(9800+.2)=ma so the acceleration was -5.39 m/s^2 for the whole system. Plugged that into Vf=Vo+at to get 1.86 seconds and plugged both into the second equation to get 9.27m for ΔX.

That wasn’t any of the possible answer choices, any help is appreciated.

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