@cpereznj.... Others have noted part of the problem, namely: in general, the average of the whole does not equal the sum of the averages of the parts.

But I believe your data reflects a number of mistakes, which compounds that fundamental problem.

The root cause of the disparity between the 5:00:00 and 5:04:19 total times is the data itself.

In original data, you assume that the speeds are accurate to the integer, and you derive the times from distance and speed of each leg.

Since that assumption is probably incorrect, the derived times are incorrect. Consequently, their sum is incorrect.

Usually, we know the distance and time, and we derive the speed. This is demonstrated below.

You can see the formulas by clicking on the cells.

The times are hypothetical. I prorated your original derived times so that their sum is 5 hours, which I assume is known to be correct (J21).

Even the distances might be inaccurate as displayed, rounded to 2 decimal places. But I assume they are accurate "enough", since they do sum to 100 miles, which again I assume is known to be correct (H21).

As you can see, the speeds in I3:I13 are not integers. But note that they round to the integers in your original data. (Truth be told, that is just a coincidence.)

(Conversely, if you want to derive times that are "accurate enough", you need to know the speeds with greater accuracy.)

Now we can talk about your arithmetic errors.

As jasonb75 wrote, the actual average speed is total distance divided by total time. That is demonstrated by the formula in I14.

In general, that is not the same as a simple average of the speeds for each leg of the trip. That is demonstrated by the formula in I15.

Instead, as jasonb75 wrote, the average speed can also be calculated by a weighted average of the speeds for each leg.

But weighted by what?

The answer is: weighted by time. That is demonstrated by the formula in I16.

We cannot weight by distance. That is demonstrated by the incorrect result of the formula in I17.

The reason that we weight by time is, again: average speed is total distance divided by total time. If you work out the algebra of the weighted average in I16, you will see that speed (distance divided by time) times time results in distance. Thus, the numerator is the sum of the distances, and the denominator is the sum of the times.

As toadstool wrote, the one exception is: if each leg took the same amount of time, the simple average of the speeds would be the same as the weighted average of the speeds.

In my example, you can demonstrate that by replacing the times in J3:J13 with the formula =$J$21/11 .

(Of course, that changes the average speed of each leg.)