Question

A car is running with a velocity 72 kmph on a level road,is stopping after travelling a distance of 30m after disengaging its engine(g=10ms/2).The coefficient of friction between the road and the tyres is

Answer 1

Answer:

2/3

Explanation:

Braking distance formula is given by

D = u²/(2μg)

From above

μ = u²/(2Dg)

= (72 × 5/18 m/s²)² / (2 × 30 m × 10 m/s²)

= 400 / 600

= 2/3