Consider a car moving on a straight road with a speed of `100m//s`. The distance at which car can be stopped is `[mu_k=0.5]`

Options

A. `800 m`
B. `1000 m`
C. `100 m`
D. `400 m`

Answer

B

Explanation

Retardation due to friction `=mug`
we know `v^(2) =u^(2) + 2as`
`:. 0 = (100)^(2) -2 (mug)s`
or `2mugs =100 xx 100`
or `s = (100xx 100)/(2xx 0.5 xx 10) =1000m` .