JEE Main Physics question
If the coefficient of static friction between the tyres and road is 0.5, what is the shortest distance in which an automobile can be stopped when travelling at 72 km h⁻¹?
Answer: (C)
Solution
v = 72 km/h = 20 ms⁻¹; a = μg = 0.5×9.8 = 4.9 ms⁻². s = v²/2a = 400/9.8 ≈ 40.8 m.