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?

Answer 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
so
24 = (Vi/2)(Vi/6.87)
Vi^2 = 329.8
Vi = 18.2 m/s

Answer 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.