After brake lock, the skidding begins.
The skidding gives a net deceleration of a = ng
dry road a = 0.84 x 9.8 = 8.232 m/s^2
wet road a = 0.58 x 9.8 = 5.684 m/s^2
using third equation v^2 = u^2 + 2as
s = u^2 / 2a
s is inversely proportional to a
s1/s2 = a2/a1
s wet / s dry = a dry / a wet = 8.232 / 5.684 = 1.448 times = 144.8 %
A driver makes an emergency stop and inadvertently locks up the brakes of the car, which...
A driver makes an emergency stop and inadvertently locks up the brakes of the car, which sklds to a stop on dry concrete. Conslder the effect of rain on this scenario. If the coefficlents of kinetic friction for rubber on dry and wet concrete are μκ(dry) = 0.84 and μκ(wet) = 0.58, how much farther would the car skid (expressed in percentage of the dry-weather skid) if the concrete were instead wet? Axwes 448276 Axdry