5
$\begingroup$

Speed is usually defined as the magnitude of (instantaneous) velocity. So one could assume that average speed would be defined as the magnitude of average velocity. But instead it is defined as

$$s_{\textrm{average}} = \frac{\textrm{total distance traveled}}{\textrm{total time needed}}$$

which generally speaking is not equal to the magnitude of the corresponding average velocity.

What historical, technical or didactic reasons are there to define average speed this way instead of as the magnitude of average velocity?

Improve this question
$\endgroup$
6
  • $\begingroup$ What are you trying to do? If you intend to add several velocities together, then divide by the number of velocities to arrive at an average velocity, you will find that this is an incorrect method. $\endgroup$
    – David White
    Sep 2, 2018 at 20:28
  • 4
    $\begingroup$ As an aside, the average velocity of any given air molecule in a still room is 500m/s. This is useful if you want to know about air pressure, but not if you care about wind speed. $\endgroup$
    – Turksarama
    Sep 2, 2018 at 22:58
  • 16
    $\begingroup$ I drive around in a circle at a constant speed of 50km/h. What is my average speed? $\endgroup$
    – user253751
    Sep 3, 2018 at 4:20
  • $\begingroup$ It is however the weighted average, where the weight is the amount of time spent at each speed. $\endgroup$
    – csiz
    Sep 3, 2018 at 12:12
  • $\begingroup$ @Turksarama: That's the average speed. The average velocity is essentially 0 in a still room, or equal to wind velocity when there's wind. $\endgroup$
    – Meni Rosenfeld
    Sep 3, 2018 at 15:23

8 Answers 8

Reset to default
43
$\begingroup$

People already answered your question from a usefulness standpoint, but I just want to add that your reasoning isn't correct:

Speed is usually defined as the magnitude of (instantaneous) velocity. So one could assume that average speed would be defined as the magnitude of average velocity.

That's not how it works. If we have

[speed] = [magnitude of velocity]

then logic dictates that we should have

average [speed] = average [magnitude of velocity]

and not

average [speed] = magnitude of [average velocity]

and, indeed, this is what we have.

Improve this answer
$\endgroup$
1
  • 5
    $\begingroup$ The technical phrasing being that average and magnitude do not commute, as Meni Rosenfeld said. $\endgroup$
    – Davis Herring
    Sep 4, 2018 at 6:48
16
$\begingroup$

Given a velocity as a function of time, the speed as a function of time is the magnitude of the velocity at each point in time. The average speed is then the average of this magnitude, as it would be for any function of time - such as density or temperature. The question of whether the average magnitude of the velocity is equal to the magnitude of the average of the velocity then becomes a conjecture to check. Since an object can move around at high speed while returning to the same place, and so have an zero average velocity with a high average speed, this shows by counter example that the average speed, defined just like any other average, is actually not equal to the magnitude of the average velocity.

It is very common to find that the average of some function is not the function of the average. For example, the average $x^2$ is not typically the square of the average of $x$.

Improve this answer
$\endgroup$
8
$\begingroup$

You can compute the mean value of the velocity vector.

However, it turns out to be useless sometimes. A trivial example is a circular motion.

The mean velocity of a full loop in a circular motion is $\vec{0}$, as velocity is pointing in one direction at first, and $\pi$ radians later it's pointing in the opposite one, so their contributions cancel out. So the average "velocity" is $\vec{0}$.

Nevertheless, this is not giving us much information. In contrast, the ratio of "circumference" to "time elapsed" gives us the actual "mean speed".

Sometimes, anyways, it can be useful to give them both. The more information, the better.

Improve this answer
$\endgroup$
7
$\begingroup$

The simple reason is that velocity can go negative, and this will affect the average. The clearest example of the difference would be a pendulum (or any other resonating system).

A pendulum swings backwards and forwards. It follows its track in one direction, accelerating, then decelerating to a momentary stop; and then reverses direction to repeat the exact same trajectory in reverse.

The pendulum follows the same path each time, in opposite directions. Since velocity is signed, the average velocity for going one way is exactly equal in magnitude to and the opposite sign to the average velocity going the other way. The average velocity is therefore zero.

The average speed of course will not be zero. It will be equal to the magnitude of the average velocity for one half of the pendulum's trajectory, because both halves have the same magnitude of velocity.

Improve this answer
$\endgroup$
5
  • $\begingroup$ This is exactly how I would have answered the question. A circular orbit with a constant speed is a similar example. $\endgroup$
    – DrSheldon
    Sep 3, 2018 at 1:09
  • $\begingroup$ Yep, true. I thought a pendulum would be easier to explain to start with, because an equal and opposite trajectory on one axis is more clearly equal and opposite. A circle needs understanding of splitting vectors into sin/cos on two axes, although I like that that is a constant speed which makes the average speed obvious. It'd make a great second example. $\endgroup$
    – Graham
    Sep 3, 2018 at 7:23
  • $\begingroup$ In the last sentence, maybe you assume that the pendulum follows a 1-dimensional path? If the pendulum has large amplitude so that we cannot neglect the vertical component of the movement, the average speed of a half-period going from one extreme angle to the opposite one, is not equal to the magnitude of the average velocity. (In my head, I consider a "fictive" particle following a horizontal straight-line path, always being directly above the actual pendulum. Its average speed will be lower, because it needs not go up and down.) $\endgroup$
    – Jeppe Stig Nielsen
    Sep 4, 2018 at 13:48
  • $\begingroup$ @JeppeStigNielsen Yes, I'm assuming a perfect non-decaying pendulum. For the OP's purposes, we can probably disregard pendulum decay. $\endgroup$
    – Graham
    Sep 4, 2018 at 14:28
  • $\begingroup$ I did not explain myself clearly enough. I am not talking about the decay. I am talking about a perfect (mathematically ideal, hence non-decaying) arbitrary-amplitude pendulum. The point mass of the idealized pendulum will not move back and forward along a straight line. It will trace out an arc of a circle. Therefore, in the circular movement, it is not true that the average speed is the magnitude of the average velocity, not even when the circular arc is traversed in only one direction (say, clockwise). $\endgroup$
    – Jeppe Stig Nielsen
    Sep 4, 2018 at 17:15
4
$\begingroup$

There error is the magnitude of velocity is not actually a definition, but is better appreciated as a consequence of the actual definition, and any materials that say it is are problematic study materials.

The actual definition of speed in both cases really is distance traveled over the time taken to travel it. It's just that in the case of instantaneous speed, the absolute value of velocity happens to be equal to the speed because the magnitude of the differential of arc length ($ds$) - i.e. the tiny increment of distance traveled - is the same as the magnitude of the differential of displacement ($d\mathbf{r}$), i.e. the vector from starting to current position. Mathematically, the correct definition of instantaneous speed is

$$\mathrm{speed} = \left| \frac{ds}{dt} \right|$$

and velocity

$$\mathrm{velocity} = \frac{d\mathbf{r}}{dt}$$

However now (for two dimensions at least) we have that $ds = \sqrt{dx^2 + dy^2}$, but also $d\mathbf{r} = dx\ \mathbf{i} + dy\ \mathbf{j}$. What is $|d\mathbf{r}|$?

Improve this answer
$\endgroup$
3
$\begingroup$

It's quite simple really. "Average speed" is the average of the speed. In general, the average of any function $f(t)$ is

$$\frac{\int_a^b f(t)\ dt}{b-a}$$

In the case of $f(t)$ being the speed $s(t)$, The integral of the speed $\int_a^b s(t)\ dt$ gives the total distance traveled, and $b-a$ is the time elapsed, resulting in the formula you mention.

Thought of another way, "Average speed" is "average magnitude of velocity", which is quite different from "magnitude of average velocity" - The modifiers "average" and "magnitude" don't commute. There is no reason to use one term when you really mean the other. In other words,

$$\int_a^b\|\mathbb{v}(t)\|\ dt \neq \left\|\int_a^b\mathbb{v}(t)\ dt\right\|$$

Improve this answer
$\endgroup$
2
$\begingroup$

As you wrote these quantities are different and give you different information so why would you want to call the magnitude of the average velocity the average speed?

If you are travelling in a car it is the average speed for the journey which you might want even if the start and finish point were the same.
What use would it be to say that the magnitude of the average velocity was zero?

Improve this answer
$\endgroup$
0
0
$\begingroup$

It is all said in the other answers. But I think it might be useful to restate things formally for a different point of view, possibly new to a physicist readership, thus requiring a more detailed explanation (apologies), and this is why I decided to make it an answer rather than a comment as I originally intended.

This is really a question of linguistics, not physics.

You have two entities/quantities of interest in your physics discourse, the average of the speed over time, and the magnitude of the average velocity, and you need to name them both. The way language works, at least in many systems of language analysis, and particularly in programming languages, which are on a level of formality that is similar to physics, is as follow (this is a gross oversimplification, at least for natural language):

You have elementary concepts that are expressed by various names or symbols, which form a vocabulary (with different categories: name, verb, scalar, function,...). Then you have syntactics rules (grammar) that you use to to build larger structures from elementary ones, such as expressions or sentences. Functional meaning is associated to syntactic rules.

Then getting the meaning of an expression or sentence is done more or less by considering that this meaning is a homomorphic image from the syntactic domain of sentence to the abstract domain of meanings. The homomorphism is defined by the correspondence between the elements of the vocabulary with elementary entities, and between the syntactic rules and their functional meaning.

This has been actually used in a very formal mathematical way to define programming languages.

This homomorphic approach, based on getting the meaning of the whole by composing (according to grammatical syntax) the meanings of the parts, is the essence, on a very trivial example, of the answer given by user541686

Now, people do not always respect syntax (just listen radio or TV). They can also take expressions and arbitrarily decide (when they have power to do so) that this meaning is other than what the homomorphism would derive. This may even happen through evolution in the case of natural language. But it just makes understanding things more complicated, possibly more prone to errors and misinterpretations.

Thus, you do not get to decide what "average speed" is. The meaning of this expression must be derived in a standard way from the functionnal meaning of average, applied to the meaning of speed, in the same way that you would for average pressure or average weight (well there is much that is implicit, and language is seldom simple).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.