Question

A car 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?

Answer 1

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