Yahoo Answers: Answers and Comments for 24 hour clock problem? [Physics]
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https://answers.yahoo.com/question/index?qid=20130206171124AAsJtSw
From Al P
enUS
Wed, 06 Feb 2013 17:11:24 +0000
3
Yahoo Answers: Answers and Comments for 24 hour clock problem? [Physics]
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https://answers.yahoo.com/question/index?qid=20130206171124AAsJtSw
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From Fred: I'm certain of exactly two things here:
1....
https://answers.yahoo.com/question/index?qid=20130206171124AAsJtSw
https://answers.yahoo.com/question/index?qid=20130206171124AAsJtSw
Thu, 07 Feb 2013 21:32:50 +0000
I'm certain of exactly two things here:
1. That you have a question in there.
2. That I can't make the foggiest sense of it.
What precisely are you trying to ask, good sir?
Is your example saying, that if 08:14:05 is chosen, the vehicle will stand at the start line until
8+14+5 = 27 s = clock time 00:00:27,
then travel 27 m/s until 08:14:05?
And that the object is to maximize the distance by judicious choice of hh:mm:ss under this rule?
(If so, I think you'd want to just choose 23:59:59, wouldn't you?)
Or are you asking for the choice of a sequence of increasing hh:mm:ss values, following that rule for each of these in turn, only starting at 00:00:00, and starting each leg at the end of the previous one, then maximizing total distance by the end? But then, it would always *end* at 23:59:59. Finding the max distance would still be interesting, however.
I'm surely missing something.
EDIT:
Hi, Al!
OK, but what ever do you mean by, 'accumulating time intervals'? How is this done/how does it happen, and who or what does it, and when?
And what effect do all these time intervals, once accumulated, have on the start time, speed, and end time of the trip?
Is it one particular time interval that the hh+mm+ss rule gets applied to, and if so, which one, and what effect do any of the other ones have on the outcome?
So you're going to go hh+mm+ss=N m/s for N s, so that 'leg' will take you N² meters. Is this repeated for each of these 'accumulated time intervals'?
When the "time starts at 00:00:00," is that when the trip starts? Or is that when this 'accumulation' phase starts? And if so, then when does the trip start?
EDIT2:
Wait, I think I'm getting it. What you're really asking is, to choose a nonnegative integer sum, N, and then, take all the integer (h,m,s) combinations that add up to that:
N = h + m + s
for which
0 ≤ h ≤ 23
0 ≤ m ≤ 59
0 ≤ s ≤ 59
and then, with n = the number of such combinations, you start at the final such time (although, what does it really matter when you start, since your total travel time and distance are both determined by N and n?) and you travel
v = N m/s for T = nN s
so that your total travel distance is
d = vT = nN² m
and the problem is to maximize d by choice of N.
Is that it?
So if you chose N = 3, then you'd have (h,m,s) =
(0,0,3), (0,1,2), (0,2,1), (0,3,0), (1,0,2), (1,1,1), (1,2,0), (2,0,1), (2,1,0), (3,0,0)
n = 10, v = 3 m/s, T = 30 s,
d = 10•3² m = 90 m
?
EDIT3:
OK, then. Oh, and I've corrected my limits on m & s in EDIT2.
So it's like you've got a 3dimensional grid of integer points, boxed in by those limits on h, m, & s in EDIT2, and you slice it with planes of constant sum of those 3 coordinates:
h + m + s = N
That sum is proportional to the distance of the plane from the origin.
You multiply the square of that sum, N², by the gridpoint count, n, of the section it makes through that box, which is (approximately) proportional to the section's area, and you want to maximize that product.
My first suspicion is that the max will occur for the section that passes through
(h,m,s) = (0,59,59), for which N = 118; n = trian(24) = 24•25/2 = 300; d = nN² = 4177200
If you bump N up one unit, of course, you gain some in N², but you lose some in n:
N = 119; n = trian(23) = 23•24/2 = 276; d = nN² = 3908436
If you drop N by one, N² loses, while n gains, but n is no longer triangular; the section through the box is now triangular, but with one corner "nipped" to make an isosceles trapezoid:
N = 117; n = trian(25)  trian(1) = 324; d = nN² = 4435236
so my suspicion was in error, and we must pursue this somewhat more complicated path.
For 82 ≤ N ≤ 118, we'll have
n = trian(142N)  trian(118N) = ½[(142N)(143N)  (118N)(119N)] = 3132  24N
The thing we are maximizing is now
d = nN² = 3132N²  24N³
d' = 72(87N  N²)
h + m + s = N = 87 is the solution
00:28:59, 00:29:58, .., 00:59:28
01:27:59, .., 01:59:27
02:26:59, .., 02:59:26
. . .
23:05:59, .., 23:59:05
n = 3132  24N = 1044
That last, 23:59:05, will be the start time, with 1044 trips of 87 s each, at 87 m/s, for a total trip distance
d = 1044•87² m = 7902036 m (about 4910 mi!)

From Al P: @Yahoo Answers:
I am going to remember the fac...
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Sun, 06 Jul 2014 01:19:07 +0000
@Yahoo Answers:
I am going to remember the fact that you deleted
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