B Average velocity as a weighted mean

[DaTario]

Hello everyone,
Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed.
My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds?

Best regards,

DaTario

 

[anuttarasammyak]

$$\bar{v}=\frac{2L}{L/v_1+L/v_2}=\frac{2v_1v_2}{v_1+v_2}$$
I am not sure what are harmonic mean and weighted average you say in this result.

 

[DaTario]

Given a set of numbers, the harmonic mean is the inverse of the artithmetic mean of the inverses of these numbers, namely,
$$ \bar v = \frac{1}{ \big( \frac{\frac{1}{v_1}+ \frac{1}{v_2}}{2}\big)} = \frac{2v_1 v_2}{v_1 + v_2}.$$

A weighted mean can be defined in this context as
$$ \bar v = \frac{w_1 v_1 + w_2 v_2}{w_1 + w_2}, $$
where $w_1$ and $w_2$ are the corresponding weights. If one takes $w_1$ as $v_2$ and $w_2$ as $v_1$ the same result is obtained. I would like to know if there is a way to offer an explanation on why this weighted mean is like this and why it works.

 

[PeroK]

I would like to know if there is a way to offer an explanation on why this weighted mean is like this and why it works.

The proportional of the total time that $v_1$ applies is $\frac{v_2}{v_1+v_2}$. This is a relative weighing of $v_2$. And vice versa.

 

[DaTario]

The proportional of the total time that $v_1$ applies is $\frac{v_2}{v_1+v_2}$. This is a relative weighing of $v_2$. And vice versa.

I guess I got it (with the help of yours). We can say that the expression
$$ \frac{v_2}{v_1 + v_2} $$
is the fraction of the total time that the car has moved with velocity $v_1$, since,
$$ T_{total} = \frac{L}{v_1} + \frac{L}{v_2}, $$
so that the fraction of the time the car has moved with velocity $v_1$ is
$$ \frac{\frac{L}{v_1}}{\frac{L}{v_1} + \frac{L}{v_2}} = \frac{1}{v_1} \left[ \frac{v_1 + v_2}{v_1 v_2}\right]^{-1} = \frac{v_2}{v_1 + v_2}$$


Thank you PeroK and anuttarasammyak.

 

[wrobel]

I would say that average of a velocity on the time interval $[t_1,t_2]$ is equal to
$\frac{1}{t_2-t_1}\int_{t_1}^{t_2}\boldsymbol v(t)dt$ provided nothing else is specified.

 

[DaTario]

I would say that average of a velocity on the time interval $[t_1,t_2]$ is equal to
$\frac{1}{t_2-t_1}\int_{t_1}^{t_2}\boldsymbol v(t)dt$ provided nothing else is specified.

Hi, wrobel.
But why is it not appropriate or correct to calculate the average velocity in this context from an integration in x (spatial coordinate) of the velocity as a function of the position, $v(x)$, defined as
$$
v(x) =
\begin{cases}
v_1, & 0 \leq x < L, \\
v_2, & L \leq x \leq 2L.
\end{cases}
$$
?

 

[wrobel]

But why is it not appropriate or correct to calculate the average velocity in this context from an integration in x

I did not say that. Eventually this story is about definitions

 

[anuttarasammyak]

v¯=2LL/v1+L/v2=2v1v2v1+v2
I am not sure what are harmonic mean and weighted average you say in this result.

Just for a fun I observe
$$ \bar{v}=\frac{\sqrt{v_1v_2}}{\frac{v_1+v_2}{2}}\sqrt{v_1v_2} \leq \sqrt{v_1v_2} $$

 

[anuttarasammyak]

Your "arismetic mean of inverse" in

Given a set of numbers, the harmonic mean is the inverse of the artithmetic mean of the inverses of these numbers, namely,

reminds me of law of synthesizede resistance of parallel ressitors
$$ \frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2} $$
They have difference of factor 2. I would like to understnd the situation difference clearly.

 

[Ibix]

Your "arismetic mean of inverse" in

reminds me of law of synthesizede resistance of parallel ressitors
$$ \frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2} $$
They have difference of factor 2. I would like to understnd the situation difference clearly.

The parallel resistors formula comes from the continuity equation relating the input current $I$ to the currents $I_i$ through the parallel resistors whose resistance is $R_i$:$$\begin{eqnarray*}
I&=&\sum_iI_i\\
\frac VR&=&\sum_i\frac{V}{R_i}\\
\frac 1R&=&\sum_i\frac{1}{R_i}
\end{eqnarray*}$$where we applied Ohm's law to each resistor and the equivalent resistance, $R$, to get from the first line to the second.

In general, the average velocity rule works by computing the total time $T$ from the sum of the times $t_i$ taken to travel distances $l_i$ at speeds $v_i$:$$\begin{eqnarray*}
T&=&\sum t_i\\
\frac{\sum_il_i}{\bar{v}}&=&\sum_i\frac{l_i}{v_i}\\
\frac{1}{\bar{v}}&=&\frac{1}{\sum_il_i}\sum_i\frac{l_i}{v_i}
\end{eqnarray*}$$which, as you see, does not have the same form as the resistance equation. Only in the case that all the $l_i$ ($i=1\ldots N$) are equal does it reduce to $$\frac{1}{\bar{v}}=\frac{1}{N}\sum_i\frac{1}{v_i}$$So the similarity is only present in the special case.

 

[anuttarasammyak]

Thanks @Ibix
So in other words
$$ \frac{1}{\bar{v}}=\sum_i \frac{r_i}{v_i} $$
where r_i is ratio of i-th segment's length to the whole length.

Voltage is shared by the parallel resistors but there is no such sharing for segment runs. Here 1/2 or 1/N comes from taking average (of inverse speed) ? What doe it mean physically? …aha it is time. Running time for unit length span.

$$ T=\int_0^T dt = \int_0^L \frac{1}{v}\ dl = L < \frac{1}{v} >_l $$
$$ L=\int_0^L dl = \int_0^T v\ dt = T < v >_t $$
Thus
$$ <v>_t = \ \frac{1}{<\frac{1}{v}>_l} $$
Furthermore
$$ <v>_l = \frac{<v^2>_t}{<v>_t} $$ for v>0 everywhere.

 

[Ibix]

What doe it mean physically?

I don't know that it really does mean anything physically. Even the similar starting points ($I=\sum I_i$, $T=\sum t_i$) come from different physical models. The current equation is a continuity condition at a junction while the time equation is the result of dividing a period. Then the derivations diverge because Ohm's law behaves differently from the definition of velocity.

I think the similarity of the equations is simply that you have two relationships that look like $x=y/z$ and two starting points that look like $X=\sum x_i$, and there are only so many ways that can shake out. In other words I think it's purely superficial mathematical similarity and doesn't grant any particular insight. But others may see something I don't.