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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.

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13y ago

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When speed is doubled the braking distance is doubled also?

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