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Grade 11IIT JEE Entrance Exam

Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (taking g = 10 m/s2)

(a) 30m

(b) 40m

(c) 12m

(d) 20m

Profile image of Simran Bhatia
12 Years agoGrade 11
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1 Answer

Profile image of Hrishant Goswami
12 Years ago

(b)

Here u = 72 km/h = 20 m/s; v = 0;

a =- mg = -0.5 × 10 =-5m/s2

As v2 = u1 + 2as,