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 28.5 m/s?

Note that the frictional force is constant and equals:

F = μ m g, where μ is the coefficient of (dynamic) friction.

So we have uniform deceleratd motion, for which we know:

v(t) = v(0) - a t
= v(0) - (F/m) t (because F = m a)
= v(0) - μ g t.

So the time it takes to reduce the speed to 0 is

t = v(0)/(μ g)

Next, the distance covered is

s(t) - s(0) = v(0) t - ½ a t²
= v(0) t - ½ μ g t²
= v(0)^2 / ( μ g) - ½ μ g v(0)^2/ (μ g)^2
= ½ v(0)^2/(μ g)

With your data:

s = ½ * (28.5 m/s)^2 / ( 0.800 * 9.81 m/s^2) = 51.7 m

51.74 m
from v^2/(2*g*r), substitute your local gravity for 9.81 to get 3-figure accuracy