A 1400 kg car moves along a horizontal road at speed v0 = 21.4 m/s. The road is wet, so the static friction coefficient between the tires and the road is only μs = 0.127 and the kinetic friction coefficient is even lower, μk = 0.0889.

Question

A 1400 kg car moves along a horizontal road at speed v0 = 21.4 m/s. The road is wet, so the static friction coefficient between the tires and the road is only μs = 0.127 and the kinetic friction coefficient is even lower, μk = 0.0889.
The acceleration of gravity is 9.8 m/s2 .
What is the shortest possible stopping dis- tance for the car under such conditions? Use g = 9.8 m/s2 and neglect the reaction time of the driver.
Answer in units of m.

Henry · Answer

Wc = M*g = 1400 * 9.8 = 13,720 N. = Normal force, Fn.

Fp = mg*sin 0 = 0. = Force parallel to
the road.

Fk = u*Fn = 0.0889 * 13,720 = 1220 N.

a = (Fp-Fk)/M = (0-1220)/1400 = -0.871 m/s^2.

Vf^2 = Vo^2 + 2a*d.
Vf = 0.
a = -0.871 m/s^2.
Vo = 21.4 m/s.
d = ?.