Acar is traveling at 50.0 mi/h on a horizontal highway. if the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

Using third equation of motion: 
 
Vf^2 - Vo^2 = 2as 

Where,
Vf = 0
Vo = 48 mi/hour
     = 0.4 ft/sec. 
a = acceleration 
s = distance 

Substituting values:

0 - 70.4^2 = 2as 

s = -70.4^2/2s 

s = - 2478.08/a (1)

Using Newton's 2nd Law of Motion
F = ma 

F = coefficient of friction x Normal force 
Normal force = mg
F = 0.103(mg) 
where 

m = mass 
g = acceleration due to gravity
   = 32.2 ftm/sec^2

Substituting values, 

0.103(m)(32.2) = ma 

Simplifying:

0.103(32.2) = a 
 
a = 3.32 ft/sec^2
Now,
Substituting the values in equation (1)
we get,

s = 2478.08/3.32 

s = 747.18 feet 

Sorry i don’t know what to do