# If the coefficient of kinetic friction between tires and dry pavement is 0.800, what is the shortest distance?

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 26.2m/s ?

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.)

• Accelerating force = (9.8 x 0.8) = 7.84N.

Acceleration = (f/m) = 7.84/1 = -7.84m/sec^2.

Shortest distance = (v^2/2a) = 43.78 metres.

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 x 0.25) = force of 2.45N. to stop.

Acceleration = (f/m) = 2.45/1 =-2.45m/sec^2.

Max. V to stop in 43.78 metres = sqrt.(2ad) = 14.65m/sec.

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