Question

A 4000 lb. car traveling at 80 mph on a level road locks its wheels and decelerates at a constant rate. It slides 580 ft. before it stops. Ignore the perception-reaction time of the driver.

Answer

A.) the time required to stop?

B.) acceleration during braking?

C.) frictional force between the tires and the road?

D.) Coefficient of friction between the tires and the road?

Answer 1

Given is:-

mass of the car M=4000lb or 1814.37kg

Initial speed of the car u=80mph or 35.7632 m/s

Distance travelled by car to stop is s=580ft or 176.784m

Now,

According to the third equation of motion

by plugging all the values we get

which gives

negative sign shows that it is declearation

Part-a

The time required to stop the car is

Part-b

acceleration during breaking is 3.62 m/s²

Part-c

The fricitonal force between the tires and the road is

Part-d

The coefficient of friction is

where N is the normal force

thus