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# The relationship whereby when speed increases so does the kinetic energy, help!

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1. So, I understand that V^2 is the speed, so therefore if the speed increases, so must the kinetic energy.
However, I don't get how this revision guide states things like "if you go twice as fast, the braking distance must increase by a factor of four to convert the extra KE."
I understand that statement, but what does it mean by "a factor of four". Elsewhere in the image, it says "increasing the speed by a factor of 3 increases the KE by a factor of 3^2 (=9)". It also says "If you double the speed, you double the value of v, bnut the v^2 means that the KE is then increased by a factor of four".

What does the "factor of four" mean and the factor of three? Please can you explain it to me?

2. (Original post by blobbybill)
So, I understand that V^2 is the speed, so therefore if the speed increases, so must the kinetic energy.
However, I don't get how this revision guide states things like "if you go twice as fast, the braking distance must increase by a factor of four to convert the extra KE."
I understand that statement, but what does it mean by "a factor of four". Elsewhere in the image, it says "increasing the speed by a factor of 3 increases the KE by a factor of 3^2 (=9)". It also says "If you double the speed, you double the value of v, bnut the v^2 means that the KE is then increased by a factor of four".

What does the "factor of four" mean and the factor of three? Please can you explain it to me?

multiplied by 4 or multiplied by 3

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