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Background

I have been running tests which involve timing an object moving a certain (fixed) distance, $s$.

Each test has been repeated 3 times, and the 3 times ($t{_{1}}, t{_{2}}, t{_{3}}$) for the object to travel $s$ m are recorded.

However, when calculating average velocity across $s$ for the 3 repeats, $\overline{v}$, I have a problem.

Problem

I can see two ways to calculate $\overline{v}$:

Method 1. Calculate the average time, $\overline{t} = \frac{t_{1}+t_{2}+t_{3}}{3}$, across the trials, then divide the distance by it: $\overline{v} = \frac{s}{\overline{t}}$

$$\overline{v}_{1} = \frac{3s}{t_{1} + t_{2} + t_{3}}$$

Method 2. Calculate the velocity for each trial, $v_{n} = \frac{s}{t_{n}}$, then calculate the average of these velocities: $$\overline{v}_{2} = \frac{v_{1}+v_{2}+v_{3}}{3}$$

While both seem to follow valid logic, the two methods are evidently algebraically different, and produce different answers (often very similar, but occasionally different enough to be concerning).

Question

What is the physical difference between the two methods?

Which method is the best to use in my circumstance?


Edit

Interestingly, both methods have been supported in answers and comments, and I think this stems from a loose definition of 'Average Velocity', both on my part and intrinsically to the phrase.

CR Drost's answer beautifully addresses this with an explanation of distance- and time-averaged velocity as well as other types.

I have accepted Mindless' answer as it succinctly explains where to use each type

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  • $\begingroup$ Use the definition of velocity. Total distance divided by total time is, by definition, average velocity. $\endgroup$ – David White Aug 12 '19 at 15:05
  • $\begingroup$ Or distance? You can define an average vs. any other quantity since the averaging function is really an area under a curve. $y_{\rm average} = \frac{1}{b-a} \int \limits_{a}^{b} y(x) {\rm d}x$. So you have a choice on the independent quantity you are averaging about. $\endgroup$ – John Alexiou Aug 12 '19 at 15:11
  • $\begingroup$ @DavidWhite For repeated measurements over a fixed distance ($s$), your suggestion is equivalent to method 1. However, the average of the calculated velocites (still using distance/time) for each measurement gives a different answer, method 2 (which has been derived in more depth by Eli in his answer).To me, both seem valid, and indeed both have proponents here in comments and answers. The point of my question was to ascertain the difference between the two and validity of one or the other in a certain case. $\endgroup$ – S. Dunnim Aug 12 '19 at 15:33
  • $\begingroup$ @S.Dunnim, why would a method seem valid when it violates the definition of the variable that you are trying to calculate? $\endgroup$ – David White Aug 12 '19 at 15:57
  • $\begingroup$ @DavidWhite Your first statement seems logically flawed to me - e.g. I couldn't say "Averaging the actual weights to arrive at an overall average weight is mathematically incorrect" - this is precisely the point of averaging 3 values, to arrive at an overall average. I'm not just adding all the numerators and the denominators in method 2, I am averaging the velocity values - Eli's answer below has a nice mathematical proof as to why it is in fact valid. The reason I haven't accepted his answer is he is adamant your method is also invalid. I believe both apply in different cases - see edit $\endgroup$ – S. Dunnim Aug 12 '19 at 16:09
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It depends on what you want to measure when you use the term "mean velocity". Whether you want to calculate mean velocity of the object over its entire distance or is it more of - single experiment repeated multiple times for more accuracy of measurement.

Method 1 leads to average velocity of object as though the process of movement is in series with each other. The formula $3s/(t_1+t_2+t_3)$ essential measures average velocity of an object that has traversed distance $3s$ over the course of the entire distance traversed. (This is related to situation 1 I described in 1st paragraph)

Method 2 however is more like situation 2 where you are interested in average velocity of the object which moves across a distance "s" and the repetition you have done is between sets of measurement.

From your question it is NOT so clear which situation you are dealing with. But if this is the case of repeated measurements to reduce error then I think method 2 is correct and not method 1.

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    $\begingroup$ Thank you for the clear answer, very helpful for recognising the physical difference between the two approaches. I am in situation 2 as you described (I thought that was relatively clear in the question however will edit to be even more specific for any future viewers), and thus will be using method 2. $\endgroup$ – S. Dunnim Aug 12 '19 at 12:18
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If you divide your path to $n$ intervals with different length $\Delta s_i$, and for each interval you need $\Delta{t}_i$ time to travel,then your average velocity is:

$$\bar{v}=\frac{1}{n}\sum_{i=1}^n \frac{\Delta s_i}{\Delta t_i}\tag 1$$

if the interval length is constant $\Delta s_i=\Delta s$ then you get for equation (1)

$$\bar{v}_2=\frac{\Delta s}{n}\sum_{i=1}^n \frac{1}{\Delta t_i}\tag 2$$

this is your result method 2, which is correct the average velocity.

in method 1 you calculate first the average time:

$\bar t=\frac{1}{n}\sum_{i=1}^n \Delta t_i$

then the average velocity

$$\bar{v}_1=\frac{\sum_{i=1}^n \Delta s_i}{\bar t}=n\,\frac{\sum_{i=1}^n \Delta s_i}{\sum_{i=1}^n \Delta t_i}=\frac{n\,s}{\sum_{i=1}^n \Delta t_i}\tag 3$$

if you compare $\bar v_2$ with $\bar v_1$ you see that the average velocity $\bar v_1$ is wrong

Edit

remarks to equation (1)

with $\frac{ds}{dt}=v $

or for discrete $v_i=\frac{\Delta s_i}{\Delta t_i} $

so the average velocity $\bar v$

$$\boxed{\bar{v}=\frac{1}{n}\sum_{i=1}^n\frac{\Delta s_i}{\Delta t_i}}$$

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  • $\begingroup$ Thanks for this answer, I'm glad you've included some clear, concise maths. Firstly, though, I'm assuming regarding your last statement, referring to "compare $\overline{v}_{1}$ with $\overline{v}_{2}$", you intended to label equation (2) with $\overline{v}_{2}$ and not $\overline{v}_{1}$? ----- Secondly, could you possibly elaborate as to why equation (3) is wrong in physical terms? I am unsure why the logic behind my method 1/your equation 3 isn't sufficient, and the argument (3) $\neq$ (2) hasn't wholly convinced me - couldn't you use the same logic in reverse? $\endgroup$ – S. Dunnim Aug 12 '19 at 14:30
  • $\begingroup$ @S.Dunnim see new version . the average velocity is unique , no other equations are valid $\endgroup$ – Eli Aug 12 '19 at 14:48
  • $\begingroup$ Once again, I really appreciate the maths, and I'm convinced this answer/ method 2 applies in my case (repeats of a test over one distance). However you claim method 1 is invalid in all cases, which you haven't convinced me of. Adrian Howard suggests in his comment that method 1 is valid and no others are ($v = \tfrac{\sum s_{i}}{\sum t_{i}} = \tfrac{s}{\overline{t}$ when $s$ is constant). Mindless (who I agree with) suggests both methods are valid and applicable in different cases. (The other case being average velocity over the total distance ) $\endgroup$ – S. Dunnim Aug 12 '19 at 15:25
  • $\begingroup$ The equation should read $v = \tfrac{\sum s_{i}}{\sum t_{i}} = \tfrac{s}{\overline{t}}$ $\endgroup$ – S. Dunnim Aug 12 '19 at 15:52
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There is a famous puzzle where you tell someone that a racecar driver has to maintain an average velocity of 60mph over three heats of some number of laps around a track in order to qualify to be in a race. This is not usually a problem because their cars typically go 80-90 mph. Well in this case some racer had some minor problem in their first heat and they came in at exactly 60mph, which was fine, but then in their second heat they blew out a tire midway through and had to complete the heat without it, averaging 30mph. The question is how fast they have to go for the third heat.

Most people will answer the obvious, “that’s fine if they can really push it and get 90mph.” That is a distance-averaged velocity: if you go for an equal distance at 60, then at 30, then at 90, your average is (60 + 30 + 90)/3 = 60.

The problem is that there is another thing which might be meant, and it is usually what is meant when we say “average speed:” we mean time-averaged velocity. On this interpretation, you were given a budget of 45 minutes to do three things which we thought would take less than 15 minutes each. Your first one took 15 minutes, your second took 30, meaning that you need to complete your last one in 0 minutes: this poor racecar driver needs to complete the heat with infinite speed.

There are in fact many other spaces which you can perform an average in, often corresponding to invertible functions $f, f^{-1}$. This highlights two of them, linear space $f(x) = f^{-1}(x) = x$ and reciprocal space $f(x) = f^{-1}(x) = 1/x,$ that happen to be self-inverse. But, for example, you could also take the average in logspace $f(x) = \log x, f^{-1}(x) = \exp x$ and then you would find a geometric mean, $$\exp\left(\frac13 \log a + \frac13 \log b + \frac13 \log c\right) = \sqrt[3]{a~b~c},$$ and if you used that to perform this mean then the racer would have to complete the heat with an average speed of $60^3/(30\cdot60) = 120\text{ mph}.$ That is neither a distance- nor a time-averaged velocity. Or you could do the reverse and do the average in exp-space, and find that the racer would have to complete the heat with an average speed of $$\ln(3\cdot e^{60} - e^{30} - e^{60}) = 60 + \ln(3 - e^{-30} - 1)\approx 60 + \ln 2,$$ or about 60.7 mph.

I should say that there is a reason to prefer the former over exp-space. Those first three are “special” in that they can be rewritten with additions, multiplications, divisions, and powers: this gives them a certain “unit covariance” that exp-space does not have, where I had to explicitly commit to do the above calculation to some notion that I was measuring speeds in mph. You cannot actually take an arbitrary function of a quantity like velocity that has units; it is not a number and does not have an objective logarithm or exponent. So, secretly, when I include the dimensions I must write $$v_\text{avg} = u~\log\left(\frac{e^{v_1/u} + e^{v_2/u} + e^{v_3/u}}{3}\right),$$and the value of $u$ matters a lot, finding $v_3 = u~\log\left(3~e^{v_\text{avg}/u}-e^{v_1/u}-e^{v_2/u}\right).$ Had I not chosen $u=1\text{ mph}$ but rather $u=\text{60 mph}$ then I would have instead found something like 79.9 mph. So there is a practical reason to stray away from certain combinations of these.

With that said, that objection does not stop us from having a lot of options, for example $\sqrt{a^2/3 + b^2/3 + c^2/3},$ that are valid “means” which are not generally part of our normal discourse.

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  • $\begingroup$ Wow thanks, +1 for an excellent answer, this has completely cleared up my confusion as to why I was getting contradictory, apparently correct answers. I've left the answer by mindless accepted, as it answered the second part of my question succinctly, however I hugely appreciate this comprehensive explanation and would accept both if I could :) $\endgroup$ – S. Dunnim Aug 16 '19 at 13:51
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The average velocity of 3 trials IS the average velocity, you should not be getting different answers if you are doing everything correctly. Are you carrying out your multiplications and divisions to enough decimal points? Leaving out slight remainders could change the answers slightly. Test each equation with whole numbers that will yield whole numbers for simplicity, you should get the same answers.

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    $\begingroup$ The calculations are run through excel so significant figures are not an issue. If you read the two methods I am suggesting carefully, you will see that they are in fact algebraically different as I have shown. $\endgroup$ – S. Dunnim Aug 12 '19 at 14:07
  • $\begingroup$ @S.Dunnim: If you are getting different answers, then one of the formulas is wrong, there cannot be two different averages. Add all 3 times, add all 3 distances (even if the same), figure velocity from there, this will be the average $\endgroup$ – Adrian Howard Aug 12 '19 at 14:28
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    $\begingroup$ Therein lies the problem - the approach you have just suggested, when the distance is fixed ($s$), results in method 1 ($\tfrac{3s}{t_{1}+t_{2}+t_{3}}$); whereas calculating the 3 velocities individually and taking an average of that results in method 2 (which is the method @Eli shows in his answer). Both produce different answers but the logic for each seems valid; either your method is incorrect, or Eli's is, or, as I suspect, mindless has hit closest to the mark with subtle differences between the two suggesting both are valid in different circumstances. $\endgroup$ – S. Dunnim Aug 12 '19 at 14:45
  • $\begingroup$ @S.Dunnim velocity should equal time divided by distance, not distance divided by time $\endgroup$ – Adrian Howard Aug 12 '19 at 14:49
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    $\begingroup$ @S.Dunnim sorry, you are right, I shouldn't stay up all night and do this $\endgroup$ – Adrian Howard Aug 12 '19 at 15:00
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Your question is not clear in particular part ,wether you want to find average velocity at distance s(which is nothing but the average of three experiment) or you want average velocity of the experiment where the object travel distance 3s in time interval t1,t2,t3 Which is calculate by total distance /total time that is your method 1 say i think now you will be able.to understand what is difference between them note first part is about method 2

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