Averages and average speed/instantaneous speed relation

[fog37]

TL;DR Summary

Averages, average speed, instantaneous speed

Hello,

Given a set of numbers, we can calculate their average and there are different types of averages (arithmetic, weighted, harmonic, geometric, etc.) The choice of the average depends on the situation. The average, also called mean or expectation value, is a number that can replace all the values in a particular calculation (addition, multiplication, etc.) and produce the same exact result.

My question is about average speed and instantaneous speed. Given that $speed =distance/time$ , average speed is the overall distance divided by the overall elapsed time while instantaneous speed is the distance traveled during an infinitesimally short time interval divided by the time interval itself. Is there any way to take the values of instantaneous speed assumed by an object and compute the average speed from them? If so, what kind of average would that be? The instantaneous speed of an object can vary from instant to instant.

Thanks!

 

[phinds]

Summary: Averages, average speed, instantaneous speed

Is there any way to take the values of instantaneous speed assumed by an object and compute the average speed from them?

THINK about it. What do you think?

 

[fog37]

Well, I don't really think (I am actually sure) assuming a discrete time and N instantaneous speed values, that average is just the arithmetic mean of the instantaneous speeds divided by N.

Any hint?

 

[Dale]

$$v_{avg}=\frac{1}{t_f-t_0}\int_{t_0}^{t_f}v(t)\ dt$$

 

[fog37]

Darn! Thank you!

 

[phinds]

Well, I don't really think (I am actually sure) assuming a discrete time and N instantaneous speed values, that average is just the arithmetic mean of the instantaneous speeds divided by N.

Any hint?

Sure. Draw a rectified sine wave that represents speed. Take the instantaneous speed at a peak. Does that look to be likely to be the average speed?

 

[fog37]

The formula for calculating the RMS (Root Mean Square) values for the waveform v(t) representing the instantaneous speed (which is always positive) over a time $T$ is $$v_{RMS} = \sqrt {\frac {1} {T} \int v(t) dt}$$

This is the same formula that Dale offered except that the RMS formula can have v(t) to be positive as well as negative, hence the squaring. But if $v(t) > 0$ always, the two formulas should give the same result, I believe.

 

[phinds]

instantaneous speed (which is always positive) over a time T

That doesn't even make sense. How can an INSTANTANEOUS speed be "over a time". That's a self-contradiction.

 

[fog37]

That doesn't even make sense. How can an INSTANTANEOUS speed be "over a time". That's a self-contradiction.

Sorry, that was poorly phrased... what I meant is the behavior of $v(t)$ graphically captured over a period of time #T#. The graph $v(t)$ vs #t# where we get the instantaneous speed at each different time instant...

 

[Stephen Tashi]

The formula for calculating the RMS (Root Mean Square) values for the waveform v(t) representing the instantaneous speed (which is always positive) over a time $T$ is $$v_{RMS} = \sqrt {\frac {1} {T} \int v(t) dt}$$

Did you leave out an exponent of 2 in the integrand?

the two formulas should give the same result, I believe.

Try finding a simple example where they do.

 

[jbriggs444]

$$v_{avg}=\frac{1}{t_f-t_0}\int_{t_0}^{t_f}v(t)\ dt$$

To be overly pedantic, this is a time-weighted average. One is adding up the speeds over tiny intervals, weighting each speed by the duration of the interval covered at that speed and dividing by the total elapsed time. We can see this a bit more clearly if we rewrite the formula using a uniform weighting function $w(t) = 1$$$v_{avg}=\frac{\int_{t_0}^{t_f}\vec{v(t)}w(t)\ dt}{\int_{t_0}^{t_f}w(t) dt} = \frac{<\text{total displacement>}}{<\text{elapsed time}>}$$It is actually an average velocity (sum of vectors) rather than an average speed (sum of unsigned magnitudes), but I will not belabor that detail. The pedant point I am aiming at is the notion of weighted averages, how a plain old "default" average is just the default case of a weighted average with a uniform weight and how the default uniform weighting can shift in its effect if we change our chosen parameterization.

As long as we are parameterizing an object's trajectory as a function of time, we naturally assume a weighting in which all times for the duration of the trajectory are weighted equally. So $w(t) = 1$ is a natural choice. Other choices for either parameterization or weighting are possible. One could, for instance, retain the parameterization over time and choose a distance-weighted average where $w(t) = |v(t)|$ and get:$$v_{avg}=\frac{\int_{t_0}^{t_f}\ \vec{v(t)}\ |v(t)|\ dt}{\int_{t_0}^{t_f}|v(t)| \ dt} = \frac{\int_{t_0}^{t_f}\ \vec{v(t)}\ |v(t)|\ dt}{<\text{trajectory length}>}$$Or we could parameterize by distance $s$ along the trajectory and use the uniform weighting and get a numerically identical result:$$v_{avg} = \frac{\int_{s_0}^{s_f}\ \vec{v(s)}\ ds}{\int_{s_0}^{s_f} \ ds} = \frac{\int_{s_0}^{s_f}\ \vec{v(s)}\ ds}{s_f-s_0}$$We could even parameterize by distance and weight by time ($w(s) = \frac{1}{|v(s)|}$) to obtain the time-weighted average. If we avoid stops:$$v_{avg} = \frac{\int_{s_0}^{s_f}\ \frac{\vec{v(s)}}{|v(s)|}\ ds}{\int_{s_0}^{s_f} \frac{1}{|v(s)|}\ ds} = \frac{<\text{total displacement>}}{<\text{elapsed time}>}$$