^Incorrect.
In order to to achieve the figure of 250km altitude in 5 minutes resulting in a 3000 km/h average rate of travel, the velocities contained between the 0 and the 16,000 over the five minutes would average out to 3000 km/h.
They cannot, as (final velocity + initial velocity)/2 = average velocity.
Hallelujah…finally, some maths so we can actually understand what it is that you don’t understand.
That statement above is simply incorrect. It would work as a calculation if and only if the velocity profile was linear which, as you’ve agreed previously, it isn’t - the fuel burns down, so mass reduces, but thrust is constant.
Average velocity is not necessarily the average of the start and end velocity.
It’s easy to prove that to yourself - just consider my 30mph car example. Your calculation would have the average velocity as the average of 30 and, say, 100mph even though the car only did 100mph for a few seconds after an hour at 30mph. Average speed is clearly not 65mph, as per your maths.
Or imagine a situation where start and end velocity are the same, but there’s a burst of increased velocity in the middle somewhere. Again, it wouldn’t work.
For general cases you need the area under the velocity time graph. If the slope is linear, your equation works - the area under a triangle is half the base times the height, which is, if you think about it, exactly what your equation is. But our graph here isn’t linear - it’s exponential, so you can’t calculate that simply.
Are we finally in agreement in this point at least?