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
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3
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
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