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Andrew Mason
#19
Jun25-07, 03:03 PM
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Quote Quote by nrqed View Post
Just out of curiosity: what definition of average do you use? If one uses a time average [itex] = \frac{1}{T} \int_0^T v_x dt [/itex] it does nto seem to me that the average will always be v/2.
Well, you are right, generally.

But consider this when the mass is subject to only deceleration. For a reduction in speed from v0 to v1 over an arbitrarily short time interval the change in momentum is m(v1-v0) and the average speed is arbitrarily close to (v1+v0)/2. If you combine n of these intervals:

[tex]E = \sum_{j=0}^n m(v_{j+1} - v_j)(v_{j+1} + v_j)/2 = \frac{m}{2}\sum_{j=0}^n (v_j^2 - v_{j+1}^2) = = \frac{m}{2}\left((v_0^2 - v_{1}^2) + (v_1^2 - v_2^2).... + (v_{n-2}^2 - v_{n-1}^2) + (v_{n-1}^2 - v_n^2)\right)[/tex]

But the intermediate terms all cancel out and you are left with:

[tex]E = m(v_0^2-v_{n}^2)/2 = m(v_n - v_0)(v_n + v0)/2 = \Delta p(v_0+v_n)/2[/tex]

If v_n = 0 then:

[tex]W = m(v_0)(v_0/2) = \Delta p(v_0)/2[/tex]

So, to be perfectly correct, one should say that the work done is the change in momentum multiplied by the average of the beginning and end speeds ([itex]W = \Delta p (v_i + v_f)/2[/itex]).

AM