Here, you want to skip to the end part where g=approx 3.7m/s2, ignoring all the values in between.
No, not at all. As I said, and as laid out nicely in the site I linked to, you have to do some calculus - some integration to find the area under the velocity time graph essentially.
As we've discussed a post or two ago, g reduces with increasing altitude, so it takes a lot longer to decelerate than 12 minutes. All covered off in that website I linked to. If you disagree, fine, but how long would it take to decelerate, and what would be the distance travelled, if g reduced in the manner generally agreed upon? You're saying it's wrong...but what's the right answer then? It can't be 12 minutes, and we know it must be bigger than that number.
I would prefer you launder your own wash.
Neatly press, fold, and drawer, or hang in the closet as you wish.
No idea what you're on about here. I've shown you calculations, websites etc. You've just said 'it isn't so', without providing any explanation.
I am telling you the missile travelling at 16,000km/h at an altitude of 250km, under no engine power, subject to g slightly above 9m/s2, will not climb an additional 4250km.
Period.
But it's not subject to g at slightly above 9ms
-2, it's subject to a g profile that reduces from just above 9 down to below 4 at apogee.
You believe in this stuff.
Not so much 'believe' as 'find the evidence for it compelling'
I do not because as I have demonstrated in numerous posts, governments lie about this type of crap ALL the time.
It is warmongering, fearmongering crap, designed to keep a terrified populace.
Governments lie, but science is science.
If you understand what d=rtmeans, then you know that 0-16,000km/h over the course of 5 minutes would not average out to a final distance of 250km.
It doesn't matter whether the high end acceleration takes place in front or at the end of the run.
For simplicity, imagine a profile where its speed doubled every minute, finishing at 16,000, and to keep the maths simple, just keep the velocity flat over each minute and then step it up at the end of each minute. So zero, then 1000km/h after one minute, then 2000km/h after 2 minutes, then 4000km/h after 3 minutes, then 8000km/h after 4 minutes, and a snap to 16,000km/h at the finish line.
Total distance in that case would be 1000/60 + 2000/60 + 4000/60 + 8000/60 = 250km
Now, clearly that's an inaccurate model, because the velocity can't just instantly double, and in the rocket example the mass is never zero - it decays, presumably linearly if the thrust is constant, to whatever the zero-fuel mass of the system is. But by reducing the time step, you can increase the accuracy, and it nevertheless illustrates the point.
I really don't get how to make this any clearer, but then it's not helped by your continued refusal to actually show any maths of your own.