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.

Re: geometric mean and velocities
 

That should be


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


Screwed up the punch line.

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.