Averaging average velocities

If the time intervals are equal in each of the three cases then you can take a mean of the means. If not, then you need to calculate total distance over total time or the three measurements don't get equal weighting.
Imagine that there is a wide variation in instantaneous velocity and that you do one measurement over a long interval and two others over very short intervals. The single average over a long interval will have reduced the error considerably but the two extra measurements could be well adrift and they could distort the overall answer out of all proportion with their significance if you take a mean of the means..

 

Gersty
The two methods give different results - one is correct, and one is not, depending upon the setup.

If you are measuring the time to travel a set distance, then the trial times are averaged before calculating the average velocity from the distance / time(avg). The average of the velocities of the trials gives the incorrect result.

If you are measuring the distance travelled in a set time, then the trial distances are averaged before calculating the average velocity from distance(avg) / time. Or in this case you could calculate each trial velocity and average all the trials.

The problem is often asked as follows.
A car travels 60 mph for 1 hour and 30 mph for one hour, what is the average velocity?
Ans, 45 mph. [ 60 miles + 30 miles ) / 2 hours ].
. Or [ ( 60 mph + 30 mph ) / 2 = 45 mph ] with velocity averages.

Which is not the same as; ( this is your experiment )
A car travels 60 miles in one hour and 60 miles in 2 hours, what is the average velocity?
Ans: 40 mph - total distance / total time
. ie [ (60 miles + 60 miles ) / 3 hours ].
Or : 40 mph - by averaging velocities for equal set time intervals
. ie .60, 30, and 30 mph each for 1 hour Avg = ( 60+30+30 ) / 3 = 40 mph
Or : 40 mph - set distance / average time
. ie avg time = 1 hour + 2 hour = 1.5 hour
. Set distance ( 60 miles ) / avg time ( 1.5 hours ) = 40 mph