JEE Main Physics question
Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of kinetic friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is [g = 10 ms⁻²]
Answer: (B)
Solution
v = 72 km/h = 20 m/s. Shortest stopping distance = v²/(2μg) = 400/(2×0.5×10) = 400/10 = 40 m.