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Preface

The definition of average speed of an object is defined by the distance travelled over time: $$v_{avg} = \frac{x_2 - x_1}{t_2 - t_1}$$ The interpretation of speed is that when you multiply speed with the time interval, you should get the distance you travelled at this interval. However, it does not measure the local variation so a better definition is the instantaneous speed: $$v = \lim_{\Delta t\rightarrow0} \frac{\Delta x}{\Delta t}$$ The above are very standard stuff for introduction mechanics course.

Average speed with equal segments

Now, if we have speed for consecutive paths with equal length, beginner students often commit an error. They may use the following: $$v_{avg} = \frac{1}{2}(v_1 + v_2)$$ as average speed between the start point and ending point, which is WRONG in general. The correct way to obtain the effective average speed is $$v_{avg}=\left(\frac{1}{N}\sum_{i}^{n}v_{i}^{-1}\right)^{-1}$$

Mean of varying speed measurements for the same path

Now, suppose that there is a situation that the distance between a starting point and an ending point is fixed. Now, there are experiments to measure the time spent by the traveller (or vehicle, or particle) moving along this path multiple time. Because the time it takes is always different, so we can obtain a list of speed $v_1, v_2, v_3, ..., v_n$ that differ with each other. Presumably there is a corresponding speed distribution $\mathcal{P}(v)$ for the speed travelling along this path. The mean of the speed distribution is given by:

$$\left\langle v\right\rangle = \int_0^\infty \mathcal{P}(v) dv \approx \frac{1}{N} \sum_i^n v_i$$

Note that it is different from the average speed that defined at the first part. It is the mean of the speed distribution function and it is similar to the "wrong $v_{avg}$" discussed in the previous part. Please do not mix it up with the average speed defined above.

So my question: I am seeking an interpretation of this quantity $\left\langle v\right\rangle$, that is, I want to know the situations/problems that this idea of $\left\langle v\right\rangle$ can be applied.

Themis
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unsym
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    This question doesn't really make sense. At the very least, it's a bit basic for this site. – Noldorin Dec 16 '10 at 16:36
  • This question could make some sense if it was generalized to the problem of measurements and role of statistical and systematic errors and such. But like this I also see no real content in it. – Marek Dec 16 '10 at 16:47
  • @Noldorin: I have made a edit to make the problem clear, so can you open it. I am not asking for the definition of average speed and it is not a trivial question. I think you may not think about this before. – unsym Dec 16 '10 at 17:01
  • Asking a definition and its application should be appropriate in this forum. Would you consider to reopen it?http://meta.physics.stackexchange.com/questions/241/should-we-allow-the-definition-and-application-questions – unsym Dec 16 '10 at 18:10
  • I have an answer to the question. It's a physics answer and it would be nice if I were allowed to post it. (hint: a bacterium with a limited amount of energy to spend swimming would be interested in this "mean of the speed distribution" quantity, if a bacterium were capable of being interested in things) – Mark Eichenlaub Dec 16 '10 at 19:39
  • I still don't understand the question. You just defined three different notions of average speed that are all fine in certain context but have absolutely nothing to do with each other. So why are you mixing them up in one question if you are really interested only in the last one? And for that last one: my first comment about measurements applies. – Marek Dec 16 '10 at 20:01
  • @Marek: I am asking for the mean of speed distribution and want people differentiate all these notation in their answers. – unsym Dec 16 '10 at 20:07
  • I have reopened this, but because it was fixed -- I agree with Noldorin's opinion about the original form. –  Dec 16 '10 at 20:37
  • @hwlau, @mbq: Yeah, it looks quite alright in its current form. (I'm still not sure I fully understand, but hopefully someone will.) – Noldorin Dec 16 '10 at 21:00
  • The harmonic mean does not actually give the average speed... – David Z Dec 16 '10 at 21:36
  • @David In the scenario described by hwlau, the harmonic mean calculates the average speed. He's describing a situation in which you travel $N$ segments, all with equal length, at different speeds. See http://physics.stackexchange.com/questions/29/how-to-calculate-average-speed – Mark Eichenlaub Dec 16 '10 at 22:08
  • @Mark: Ah, I missed the part about equal length. That would be the special case in which it does work. – David Z Dec 16 '10 at 22:26
  • Why is the average velocity $v_{avg}=\left(\frac{1}{N}\sum_{i}^{n}v_{i}^{-1}\right)^{-1}$? I don't understand why we need both the ^-1 – raindrop Dec 27 '12 at 21:39
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    @Raindrop This is called harmonic mean, it is used to keep the displacement divided by time meaningful. For more detail, you can check it up in wiki, or look for the accepted answer in quesntiomn http://physics.stackexchange.com/questions/29/how-to-calculate-average-speed?rq=1 – unsym Dec 27 '12 at 21:46

3 Answers3

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Suppose an object travels through a fluid at low Reynolds number. Then the drag force on it is proportional to its velocity. In an talk called "Life at Low Reynolds Number", Ed Purcell wrote about bacteria swimming at low Reynolds number,

For these animals inertia is totally irrelevant. We know that F=ma, but they could scarcely care less. I'll show you a picture of the real animals in a bit but we are going to be taking about objects which are the order of a micron in size... In water where the kinematic viscosity is $10^{-2}$ cm/sec these things move around with a typical speed of $30$ micron/sec. If I have to push that animal to move it, and suddenly I stop pushing, how far will it coast before it slows down? The answer is, about $0.1$ angstrom. And it takes it about $0.6$ microsec to slow down. I think this makes it clear what low Reynolds number means. Inertial plays no role whatsoever.

Thus, if you're a bacterium, then energy you need to use to take a stroll can be calculated solely from the energy needed to overcome drag. The drag is proportional to velocity, so the energy needed to travel a certain distance is proportional to the velocity you travel at while doing it.

The mean of the speed distribution, defined by

$$\langle v \rangle = \int_{path} v(x) dx$$

is proportional to the energy expended per unit distance traveled.

Mark Eichenlaub
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    I may be missing something but this looks completely different than what hwlau asked for. He asked for statistical distribution of speed measurement (which would presumably be gaussian around the mean). You are talking about distribution of the speed in space. On the other hand, I am not sure there exists any reasonable answer to the hwlau's question in statistical sense (which was my complaint in the comments above) so maybe it's good that you generalized it and provided unrelated but interesting answer :-) – Marek Dec 16 '10 at 23:08
  • @Marek I was going off this: "if we have speed for consecutive paths with equal length". I.E. my interpretation was that hwlau was saying that we traverse many equal-length distances at different speeds. We write down a lists of the speeds for each distance, then take the average of the entries on the list. I was led to this by looking at the equation that sums the speeds and divides the by number of entries. Maybe I was barking up the wrong tree, though. – Mark Eichenlaub Dec 16 '10 at 23:18
  • interesting interpretation; you combined harmonic average (which is where the part about the equal lengths comes from) and distribution mean in a way I could never have dreamed of :-) – Marek Dec 16 '10 at 23:47
  • Not exactly my question. But you answer gives an interesting point about the speed and force. And your answer only correct when the interval tends to zero. It is speed in the question, so it may take a longer path and spend more energy for each interval. – unsym Dec 17 '10 at 12:00
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The usual (and correct) method is to take the average speed over time. What you are defining is the average speed over distance. In computing average speeds, we have to take the average over time basically because speed has time in the denominator. That is, distance is given by the product of velocity and time so in computing distances correctly we have to average over time. In other words, what we seek is proportional to time.

An area where the reverse definition would be appropriate is one where we are looking for an effect that is proportional to distance, but still depends on velocity. An example might be the wear and tear on roads, as a function of the speeds of the vehicles driving on it. Or a computation of the number of fatalities per mile. Average fuel consumption in gallons per mile might do the trick.

Carl Brannen
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In the room where you are, the gas molecules around you are moving at various speeds (scalar value). The probability distribution of speeds is $\mathcal{P}_v$. It is the probability (a fraction) of finding a molecule at a given speed value.

For an infinite distribution of speeds, the mean speed of a gas molecule in the room is found by taking the probability times speed and integrating over all speeds from zero to infinity. We obtain the finite sum expression when we take $\mathcal{P}$ as constant for all speed values and assume speed values vary in discreet increments. For a discreet distribution, the mean is instead better termed as the average $\langle v \rangle \rightarrow v_{avg}$.

A more robust form of the expression is to use probability density $\rho_v \equiv \mathcal{P}/v$. Probability density is the probability that a molecule is at a given speed with a given infinitesimal window of speed. We then write the expression below for the mean speed.

$$\langle v \rangle = \int \rho_v(v)\ v\ dv$$

Finally, you can go do a simple experiment to appreciate the analysis. Go out to your nearest interstate or autobahn or equivalent. Sit for an hour or so and record the speed of cars that pass you. Bin the speeds into discreet increments, perhaps blocks of $5$~mph or kph. Note the number of cars within each bin. Divide by the total number of cars that you measured. You have created a distribution of probability $\mathcal{P}$ (fraction of cars at a given speed). Divide the values again by the middle speed at the bin. You have translated the probability distribution to a probability density distribution.

Jeffrey J Weimer
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