Question

(a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop a car by locking the brakes when the car is traveling at 28.7 m/s (about 65 mi/h)? (b) On wet pavement the coefficient of kinetic friction may be only 0.25. How fast should you drive on wet pavement to be able to stop in the same distance as in part (a)? (Locking the brakes is not the safest way to stop.)

Answer 1

a) 52.5 m

b) 16.0 m/s

Explanation:

a)

The motion of a car slowed down by friction is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$

where

v = 0 is the final velocity (the car comes to a stop)

u = 28.7 m/s is the initial velocity of the car

a is the acceleration

s is the stopping distance

For a car acted upon the force of friction, the acceleration is given by the ratio between the force of friction and the mass of the car, so:

$a=\frac{-\mu mg}{m}=-\mu g$

where:

$\mu=0.80$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting and solving for s, we find:

$s=\frac{v^2-u^2}{-2\mu g}=\frac{0-(28.7)^2}{-2(0.80)(9.8)}=52.5 m$

b)

In this case, the car is moving on a wet road. Therefore, the coefficient of kinetic friction is

$\mu=0.25$

Here we want the stopping distance of the car to remain the same as part a), so

$s=52.5 m$

We can use again the same suvat equation:

$v^2-u^2=2as$

And since the final velocity is zero

u = 0

We can find the initial velocity of the car:

$v=\sqrt{2as}=\sqrt{2\mu gs}=\sqrt{2(0.25)(9.8)(52.5)}=16.0 m/s$