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#14
Mar17-12, 07:18 PM
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P: 14,442
You don't need kinetic energy. There's nothing special about velocity here; I would take that v to mean any arbitrary vector quantity that satisfies [itex]\mathbf v \cdot \mathbf v ' = 0[/itex]. All you just need to do is to look at the derivative of the inner product of this vector v with itself.