Question

1)If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance in which an automobile can be stopped by locking the brakes when traveling at 26.0m/s ?

2)On wet pavement the coefficient of kinetic friction may be only 0.250. How fast should you drive on wet pavement in order to be able to stop in the same distance as in part A? (Note: Locking the brakes is not the safest way to stop.)

Answer 1

1)

Force of friction f = ?N = ?mg,

       where m is the massof the car.

       Aceleration,

                  a = f/m =?mg/m = ?g = 0.8 x9.8 = 7.84m/s²

Theditance travelled, S before coming to rest (v = 0)

        S = (v²- u²)/2a

           =(0 - 26x26)/2x7.84

           =43.11 m

2)

Acceleration, a = ?g = 0.25 x9.8 =2.45 m/s²

               u² = v² - 2a.S

                  = 0 - 2 x-2.45x43.11 = +211.24 m

  Initial velocity,

               u = ?211.24 = 14.53 m/s