#1
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geometric mean and velocities
I just noticed irchans question way down below about whether there is an application of the geometric mean to speeds. I found that the geometric mean gets involved if we consider the average speed for two different accelerations. Consider accelerating from a standstill: distance = acceleration * time^2 d = at^2 accelerate over distance d twice: avg. speed = 2d/[sqrt(d/a1) + sqrt(d/a2)] = sqrt(d)sqrt(a1a1)/[sqrt(a1)+sqrt(a2)]/2 For example if we start from a standstill and accelerate at 6 mph/sec for a mile, and then do it again at 4 mph/sec, the average speed from the above formula will be 132 mph. If you plug 6 and 4 into the above forumlas, you have to multipy the result by sqrt(3600) to convert seconds to hours. |
#2
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Re: geometric mean and velocities
That should be sqrt(d)sqrt(a1a2)/[sqrt(a1)+sqrt(a2)]/2 Screwed up the punch line. |
#3
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correction
I'm missing a factor of 1/2 d = (1/2)at^2 since v = at and d = integral[(at)dt] so the final equation is: avg. speed = 2sqrt(2d)sqrt(a1a2)/[sqrt(a1)+sqrt(a2)] My example of 6 mph/sec and 4 mph/sec for a mile gives an average speed of 187 mph. |
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