demagogue on 22/11/2008 at 19:42

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Let's see, you're going to reach 0 mph long before you reach your destination. Edit: oops, I guess I'm thinking about reducing speed every minute, rather than every mile. That would change the problem, so ignore the rest. Well, it still does part of the work, but you need to add in the difference, which I do in my next post...

Think about the 60 mile distance divided evenly into 1/60 parts, each one 1 mile = 1 minute of travel @ 60 mph. As you slow down, a progressively bigger chunk is taken out of each 1/60 section that you don't travel in 1 minute (0/60, 1/60, 2/60...). Sum all of that up, and that's the distance gap between hitting 0 mph and your destination, Σ[0->59]n/60 (nevermind that the gap is in bits in pieces; the math is the same) ... which is 1770/60 = 29.5 miles gap, so you travel 60-29.5 = 30.5 miles and then stop.

As for time, each of those 1/60 chunks still takes 1 minute to travel, and since you traveled all 60 of them and then stopped (not the full length of them; but the 1 minute of travel of them), it still takes 1 hour of travel start to finish.

demagogue on 22/11/2008 at 19:53

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That's why I start the sum at 0/60. It's subtraction, and 1 mile - 0/60 = one mile, just like it should.

Edit 2: Oh, you're probably talking about this instead:

Oops, that's obviously a typo, now corrected. I meant (and actually thought I typed) 1/60 one mile parts, but I guess I got distracted and typed it out wrong. The math is still right.
Edit 3. Well almost... But what's wrong isn't that.

heywood on 23/11/2008 at 01:51

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Well, almost. Each bounce takes more than half the time of the previous bounce. In the air, the ball moves with a constant acceleration (gravity). The distance traveled by a body under constant acceleration is d = a * t^2 / 2. If it bounces half as high, the bounce takes 1/sqrt(2) as long. That produces the following sequence:

1 + 1/sqrt(2) + 1/2 + 1/(2*sqrt(2)) + 1/4 + 1/(4*sqrt(2)) + 1/8 + ...

The odd order terms (1, 1/2, 1/4, ...) form one sequence that we already know is convergent and sums to 2. The even order terms are just 1/sqrt(2) multiplied by the same sequence and therefore sum to 2/sqrt(2) = sqrt(2). So the total time it takes is 2+sqrt(2) times the initial bounce time.

TBE on 23/11/2008 at 17:10

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For my equation, it takes 1 minute to travel the first mile, then when you slow to 59 miles per hour, it would take 1.0169491525423728813559322033898 minutes to travel that mile or (0.016949152542372881355932203389831 hour), and so on. So when you're 30 miles from the target, you'd be traveling at 30 miles per hour, which would take you 2 minutes to travel that mile. I guess, time = distance/rate for each mile, then change the distance/rate for each mile you travel 1/60, 1/59, 1/58, 1/57, etc. Then store the result for each mile, and add them at the end to find the total time. So the last mile, you're going 1 mile per hour, which will take you 1 hour. You end up at the destination 0, and there are no further calculations, as you've reached the destination. I don't have visual basic or anything installed, or I'd program something myself.

demagogue on 24/11/2008 at 08:51

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It would be easier (to read anyway) if you wrote those numbers as sum fractions.

At 59 mph, you'd go 59/60 the distance of one mile in one minute. That leaves +1/60 mile @ 59 mph (or 59/60 mi/min) = [(1/60)/(59/60)]min = 60/(60*59) = yes, +0.0169 min = 1.0169 min's for that second mile, same that you got.

As a sum ...

I already computed the % of 1 minute travel per mile (http://www.ttlg.com/forums/showthread.php?p=1792187#post1792187) above, using this approach, and that was 1 hour of time (obviously) that got you 30.5 miles; so you've got the extra 29.5 miles to add up, in 60 progressively-bigger pieces, going the progressively slower speeds for each piece.

So we take that 1 hour of time and add the extra bits, turning the [(1/60)/(59/60)]min test fraction into a sum per mile:

+Σ[0->59] (n/60) / [(60-n)/60] =
+Σ[0->59] (n * 60) / [(60-n) * 60] =
The 60's cancel out
So it's going to be
1 hour + Σ[0->59] n / (60-n) minutes

Let's check a few terms, per mile...
+0/60 = 0 (first mile is exactly 1 minute, no extra added)
+1/59 = our beloved 0.0169
+2/58 = 0.0344
...
+59/1 = 59 min's (last mile is an hour; again the extra 1 minute is already part of the starting 1 hour)

Ok, looks good. Although of course, fuck, since the denominator is changing every term, there's not really any shortcut to doing the division every single stinking time. It feels like there might be a shortcut, though, since (almost) half are 1/x the other half. Also, half can have a denominator of 2. Well, screw it, I'm sure Ramanujan could figure it out, but nothing is clicking for me.

In bold is the formula to make a program churn it out in a few nano-seconds, though...