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?

$\endgroup$
  • $\begingroup$ 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 '18 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 '18 at 22:58
  • 14
    $\begingroup$ I drive around in a circle at a constant speed of 50km/h. What is my average speed? $\endgroup$ – immibis Sep 3 '18 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 '18 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 '18 at 15:23
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.

$\endgroup$
  • 4
    $\begingroup$ The technical phrasing being that average and magnitude do not commute, as Meni Rosenfeld said. $\endgroup$ – Davis Herring Sep 4 '18 at 6:48
15
$\begingroup$

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 $x$.

$\endgroup$
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.

$\endgroup$
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.

$\endgroup$
  • $\begingroup$ 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 '18 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 '18 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 '18 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 '18 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 '18 at 17:15
3
$\begingroup$

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 $|d\mathbf{r}|$?

$\endgroup$
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,

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

$\endgroup$
1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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