A driver sees a horse on the road and applies the brakes so hard that they lock and the car skids to stop in 24 m .The road is level ,and the coefficient of kinetic friction between tires and the road is 0.7.How fast was the car going when the brakes were applied?

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  1. weight = m g
    so F = -mu m g =- 0.7 * m * 9.81 =- 6.87 m
    so acceleration = - 6.87 m/s^2
    v = Vi + a t
    0 = Vi - 6.87 t
    t = Vi/6.87
    average speed = Vi/2
    24 = (Vi/2)(Vi/6.87)
    Vi^2 = 329.8
    Vi = 18.2 m/s

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  2. M*g = 9.8M = Wt. of vehicle = Normal(Fn).

    Fp = Mg*sin A = 9.8M*sin 0 = 0. = Force parallel with surface.

    Fk = u*Fn = 0.7 * 9.8M = 6.86M N. = Force of kinetic friction.

    Fp - Fk = M*a,
    0 - 6.86M = M*a,
    a = -6.86 m/s^2.

    V^2 = Vo^2 + 2a*d = 0,
    Vo^2 - 13.72*24 = 0,
    Vo = 18.1 m/s.

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