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08-14-2002, 11:56 AM
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.

08-14-2002, 12:00 PM
That should be

sqrt(d)sqrt(a1a2)/[sqrt(a1)+sqrt(a2)]/2

Screwed up the punch line.

08-14-2002, 02:17 PM
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.