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.

Answer 1

U = initial speed = 20 m / s

V = Final velocity = 0

We take g = 10 m/s²

We need to find the acceleration which will be negative given that the car is coming to rest.

A = - 0.4 × 10 = -4 m / s²

From the formula :

V² = U² + 2as

We can get S

0 = 20² - 2× 4 × S

0 = 400 - 8S

S = 400/8 = 50m

Answer 2

The answer is 50 m.

I hope solution is understood