Question

A car is traveling at 53.0 mi/h on a horizontal highway.

(a) 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?
m

(b) What is the stopping distance when the surface is dry and µ_s = 0.604?

Answer 1

Let us first Convert the speed to m/s

(53.0 mi/h)(1.609 km/mi)(1000 m/km)(1 h / 3600 s)

= 23.69 m/s

The deacceleration because of the friction will be

a=μg

a=0.1*9.81

a= 0.981

(a) now applying third equation of motion,

2 a d = vf² - vi²

d = (vf² - vi²) / (2a)

= [0 - (23.69 m/s)²] / [(2)(-0.981 m/s²)]

d = 286.04 m

(b) it is also, the same only the deacceleration will change

a=0.604*9.81

a=5.92524

So,

d=(23.69²)/(2*5.9254)

d=47.3568 meters