Calculating mean velocity: using time or velocity? 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
 A: 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.
A: 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 $\bar v$
$$\boxed{\bar{v}=\frac{1}{n}\sum_{i=1}^n\frac{\Delta s_i}{\Delta t_i}}$$
A: 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.
A: 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
A: 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.
