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| ``working out the exercise''
by akrowne on 2002-01-16 07:12:12 |
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| Let the distance between the cities be d.
The velocity 1 way is x.
The velocity back is y.
so, with rate*time=distance, we have times:
x*t_1=d
y*t_2=d
hence
t_1=d/x
t_2=d/y
the total time is simply the sum of the time each way:
t=t_1+t_2
the overall velocity, v, will still conform to rate*time=distance, except the distance is twice the distance between the cities:
v*t=2d
v*(t_1+t_2)=2d
v=2d/(t_1+t_2)
=2d/(d/x+d/y)
=2/(1/x+1/y)
Hence we have derived the harmonic mean.
In general, when going between two cities n times, with {x_1,x_2,...,x_n} the velocities for each transit, we have
v*t=n*d
v*(t_1+t_2+...+t_n)=nd
v=nd/(t_1+t_2+...+t_n)
=nd/(d/x_1+d/x_2+...+d/x_n)
=n/(1/x_1 + 1/x_2 + ... + 1/x_n)
Which is the n-value harmonic mean.
-apk |
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