Weighted Average Velocity between r1 and r2

[natski]

Consider a velocity which is a function of position r, which does not vary linearly with time.

Consider a body moving with this varying velocity between distance r1 to r2.

Let us define the average velocity between r1 and r2 as (r2-r1)/time taken to travel between r2 and r1.

I assumed the average would be found by:

Integral[ v(r) dr {r2, r1}] / Integral [dr {r2, r1}]

But this formula does not seem to work. Are there any special cases where this formula is not sufficient?

 

[Meir Achuz]

You need the integral over time to get the average velocity.

 

[natski]

But time is a function of position, so there does not seem any reason why one could not use the position as an equivalent weight?

So you suggest:

Integral[v dt] / Integral[dt]

But
r(t) => dr/dt = 1/f'
=> dt = f' dr

Hence one could write Integral[v f' dr] / Integral[f' dr]

But dr/dt = v
hence f' = 1/v
=> vbar = Integral[dr] / Integral[dr /v]

 

[Shooting Star]

So you suggest:

Integral[v dt] / Integral[dt] ...(1)

...

But dr/dt = v
hence f' = 1/v
=> vbar = Integral[dr] / Integral[dr /v] ...(2)

The two integrals (1) and (2) are identical. Sorry, I don't get your point now...?

 

[natski]

Well Meir felt you couldn't take the weighted average by integrating over position, but I am trying to prove that you can by virtue of the fact that his suggestion, of integrating over time, can be written as an integral over position.

 

[learningphysics]

Well Meir felt you couldn't take the weighted average by integrating over position, but I am trying to prove that you can by virtue of the fact that his suggestion, of integrating over time, can be written as an integral over position.

Ok. But Meir was responding to this in your original post:

"Integral[ v(r) dr {r2, r1}] / Integral [dr {r2, r1}]"

which is not going to give average velocity.

But "vbar = Integral[dr] / Integral[dr /v]" in your 3rd post will give average velocity.