A car is moving along a straight horizontal road with a speed `v_(0)` . If the coefficient of friction between the tyres and the road is `mu` , the shortest distance in which the car can be stopped is

Options

A. `(mu^(2))/(mu g)`
B. `(2 mu^(2))/(mu g )`
C. `(mu^(2))/(2 mu g )`
D. `((mu)/(mu g ))^(2)`

Answer

C

Explanation

Using `v^(2) - u^(2) `= 2as, we have
Here, v = 0
a = `mu`g
`therefore " " - mu^(2) = 2 (mu g) ` S
So, the retarding distance = `(mu^(2))/(2 mu g)`