What happen to the braking distance if the speed is doubled and why?

Hi there! Assuming that the deceleration (or negative acceleration, if you will) is constant and the same in both cases, you can use a special kinematic formula to solve the problem. The formula follows:

(final velocity)^2 = (initial velocity)^2 + [ 2 * (deceleration) * (braking distance) ]

Rearranged to our needs the formula reads:

braking distance = [1/2] * -(initial velocity)^2 / (deceleration)

* this equation assumes that the final velocity is zero

If the initial speed were doubled then the general formula would read:

braking distance = 2 * -(initial velocity)^2 / (deceleration)

NOTICE that the two equations are the exact same except for the leading coefficients. 1/2 is assocaited with the braking distance of the normal velocity while 2 is assocated with the breaking distance of the doubled velocity. Since 2 is four times larger than 1/2, this leads us to the conclusion that the breaking distance for an object traveling at double a certain velocity would be 4x greater than the breaking distance of the object moving at the "regular" velocity.