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# Why if the speed of a car is doubled its braking distance is much more than doubled?

Updated: 9/18/2023

Wiki User

11y ago

Because kinetic energy KE ~ V^2 (varies as the square of the speed). So ke ~ v^2 and KE ~ V^2 and when V = 2v, doubled, we have KE/ke = (V/v)^2 = (2v/v)%2 = 4 so that KE = 4 ke. QED. The new kinetic energy is four times the old.

And, ta da, that means there is four times as much energy for the brakes to sap and reduce to zero kinetic energy, which means V = 0 is the end speed (stopped). So by the work function, which you should know by now, we have WE = FS where F is the braking force (friction) and S is the stopping distance. We assume the braking force remains the same for both speeds.

Then KE - WE = 0, meaning the kinetic energy is sapped by the work so there is none left. And we have KE = WE = FS; so S = WE/F = KE/F and the stopping distance varies as the kinetic energy. So when the speed is doubled and the kinetic energy is quadrupled, the stopping distance is quadrupled because there is now four times as much kinetic energy to expend in stopping. QED.

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11y ago

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Q: Why if the speed of a car is doubled its braking distance is much more than doubled?
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