A 940 kg car moves along a horizontal road at speed v0 = 16.8 m/s. The road is wet, so the static friction coefficient between the tires and the road is only μs = 0.196 and the kinetic friction coefficient is even lower, μk = 0.1372.

Question

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

Answer 1

uk = a/g = 0.1372
a = 0.1372 * g = 0.1372 * (-9.8) =-1.34
m/s^2.

V^2 = Vo^2 + 2a*d = 0 @ max. ht.
d = -(Vo^2)/2a = -(16.8^2)/-1.34 = 210 m