Question

Part A: If the coefficient of kinetic friction between tires and dry pavement is 0.81, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 30.6 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.)

Answer 1

A) 59.3 m

The frictional force acting on the car is

$F_f = - \mu mg$

where

$\mu=0.81$ is the coefficient of friction

m is the mass of the car

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

According to Newton's second law, the acceleration of the car is equal to the frictional force divided by the mass, so:

$a=\frac{F_f}{m}=-\mu g=-(0.81)(9.8 m/s^2)=-7.9 m/s^2$

So we can find the distance travelled by the car using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity of the car, when it comes to a stop

u = 30.6 m/s is the initial velocity

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0-(30.6 m/s)^2}{2(-7.9 m/s^2)}=59.3 m$

B) 17.2 m/s

In this case, the coefficient of friction is

$\mu=0.25$

so the acceleration of the car is

$a=-\mu g=-(0.25)(9.8 m/s^2)=-2.5 m/s^2$

We can find the initial velocity of the car by using again

$v^2 - u^2 = 2ad$

where the stopping distance is equal to before:

d = 59.3 m

So, solving the equation for u we find

$u=\sqrt{v^2-2ad}=\sqrt{0^2-2(-2.5 m/s^2)(59.3 m)}=17.2 m/s$