Question

If the coefficient of kinetic friction between tires and dry pavement is 0.93, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 30.1 m/s ?

Part B

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 not the safest way to stop.)

v =

nothing

  m/s

Answer 1

Solution:

Let the mass be 1kg.

a) Accelerating force = 9.81*0.93 = 9.1233N.
Acceleration = (f/m) = 9.1233/1 = -9.1233m/sec^2.
Shortest distance = (v²/2a) = 49.654 metres.

b) Locking the brakes may not be the safest way to stop, but you are not given a static friction coefficient, unless the 0.25 is meant to be static friction?
9.8* 0.25) = force of 2.45N. to stop.
Acceleration = (f/m) = 2.45/1 =-2.45m/sec^2.
Max. V to stop in 49.654 metres = sqrt.(2ad) = 15.6 m/sec.

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