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