consider a car moving along a straight horizontal road with the speed of 72km/hr. if the coefficientof static friction between the tyre and the road is 0.5 , the shortest distance in which the car can be stopped is (g=10m/s^2)a) 30mb)40mc)72md)20m

Let ‘m’ be the mass of the vehicle. So, frictional force experienced by the vehicle on its tyres is, f = µmg

So, the acceleration of the vehicle due to this frictional force is, a = -µg (acceleration is directed opposite to the motion of the vehicle)

Now using,

v2 = u2 + 2as

=> 0 = v2 - 2 µgs

[initial velocity = v, final velocity = 0]

So, s = v2/(2µg)

v = 72 *5/18 = 20 m/sec

Hence

S = 400 /2 * 0.5 *10 = 40 m

Regards

Arun (askIITians forum expert)