Question

If the coefficient of kinetic friction between tires and dry pavement is 0.90, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 22.9m/s ?

On wet pavement, the coefficient of kinetic friction may be only 0.25. 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 notthe safest way to stop.)

If the coefficient of kinetic friction between tires and dry pavement is 0.90, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 22.9m/s ?

Answer 1

(a) Force of friction f = ?N = ?mg,

       where m is the massof the car.

       Aceleration,

                  a = f/m =?mg/m = ?g = 0.0 x9.8 = 8.82 m/s²

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

        S = (v²- u²)/2a

           =(0 - 22.9x22.9)/2x8.82

           =59.46 m

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

               u² = v² - 2a.S

                  = 0 - 2 x-2.45x59.46 = +291.354

  Initial velocity,

               u = ?291.354 = 17. 07 m/s