When the speed of car is doubled, the distance required to stop it becomes 4 times. Why?

Explanation

Let the retardation produced by applying the brakes be a. Let its initial velocity be u and the distance covered by the car before coming to rest be s.
Hence, 0 = `u^(2) -2as` [ from equation `v^(2) = u^(2) -2as`]
or,`" " s = (u^(2))/(2a)`
If the retardation is constant then `s prop u^(2)`.
Hence if u is doubled, s becomes `2^(2)` or 4 times.