Question

A car is moving on a straight road at a speed of 20 m/s the coefficient of friction between the tyres of the car and the road is 0.4,find the shortest distance within which the car can be stopped.(g

Answer 1

Answer:

Correct option is C)

Acceleration due to friction (retardation) (a) = - μg=−0.9×10=−9m/s

2

Final velocity, v=0m/s

Initial velocity, u=20m/s

Applying formula: v

2

=u

2

+2as

Where s is the distance traveled by car before it come to rest.

⇒0

2

=20

2

+2(−9)s

⇒18s=400

⇒s=22.2 m