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Topic: Two Bees ... or Not Two Bees? (Read 16724 times) |
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rloginunix
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Two Bees ... or Not Two Bees?
« on: Jul 13th, 2016, 5:50pm » |
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Two Bees ... or Not Two Bees?
When, is the question.
At sunrise a bee B1 took off from the flower F1 and flew with constant velocity toward the flower F2 from which, at the same time (at sunrise), a bee B2 took off and flew with constant velocity toward the flower F1.
After both bees met at high noon, B1 touched down on F2 at 16:00 while B2 landed on F1 at 21:00.
At what time did the sun rise that day?
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towr
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Re: Two Bees ... or Not Two Bees?
« Reply #1 on: Jul 14th, 2016, 11:17pm » |
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By construction in Cinderella, it seems to be about 6 hours before noon.
By the law of tidy math solutions it's probably exactly that.
So let's say the distance between F1 and F2 is 30 units
Then B1 flies at a speed of 30u/10h = 3u/h and B2 at 30u/15h = 2u/h
Then after 6 hours B1 is at a distance of 3*6=18 from F1, and B2 is at a distance of 2*6=12 from F2 and thus 18 from F1.
So that confirms the solution.
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Grimbal
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Re: Two Bees ... or Not Two Bees?
« Reply #2 on: Jul 15th, 2016, 2:46am » |
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I agree. Assuming F1 and F2 are in the same time zone. 
B1 flew 4 hours after the meeting.
B2 flew 9 hours after the meeting.
The flight times before the meeting and after the meeting for a bee are in a ratio matching that of the distances flown before and after the meeting. That ratio is reversed between B1 and B2.
If the flight time before the meeing is x, 4/x = x/9.
You see that x is the geometric mean between 4 and 9, thats is 6.
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| « Last Edit: Jul 15th, 2016, 2:48am by Grimbal » |
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rloginunix
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Re: Two Bees ... or Not Two Bees?
« Reply #3 on: Jul 15th, 2016, 8:50am » |
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Thank you for keeping me on my toes, Grimbal:
- F1 and F2 are in the same time zone
- |F1F2| <<< Re
- vi <<< c
towr is correct.
Grimbal is correct.
(my solution path resembled that of Grimbal's: if |F1F2|=d, M is a meeting point, |F1M|=x, then from flight time to M equality: x/v1 = (d-x)/v2. After the meeting, B1 covered B2's distance to meeting, d-x, and conversely: d-x=4v1 and x=9v2. Hence, in closed form: t=9v2/v1=4v1/v2)
Peculiar: the answer is a product of two smallest primes whose squares are used in the interim.
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alien2
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Re: Two Bees ... or Not Two Bees?
« Reply #4 on: Jul 15th, 2016, 9:24am » |
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Two bees or not two bees - that is the question:
Bees are fearsome in the minds of those who suffer,
With stings like arrows of outrageous fortune;
Should we take arms against a sea of wasps,
And, by opposing, end them?
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rmsgrey
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Re: Two Bees ... or Not Two Bees?
« Reply #5 on: Jul 15th, 2016, 12:52pm » |
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on Jul 15th, 2016, 8:50am, rloginunix wrote:| Peculiar: the answer is a product of two smallest primes whose squares are used in the interim. |
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That's really not that peculiar - the general solution, with bees meeting at t=0 and arriving at their final destinations at times t=a and t=b, has them starting at t=-SQRT(ab). In order for all times to be integers, either a=b or a,b are both perfect squares. To fit within a day, {a,b} has to be a subset of {1,4,9} (unless you measure time in smaller units than hours) and {4,9} gives the most interesting answer of the simple answers.
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JiNbOtAk
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Re: Two Bees ... or Not Two Bees?
« Reply #6 on: Aug 24th, 2016, 5:49pm » |
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on Jul 15th, 2016, 2:46am, Grimbal wrote:The flight times before the meeting and after the meeting for a bee are in a ratio matching that of the distances flown before and after the meeting. That ratio is reversed between B1 and B2.
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I remember facing this type of questions during my high school physics. If only my teacher had used Grimbal's approach, it would had been far easier to understand (rather than brute-forcing through with the mathematical approach) 
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alien2
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Re: Two Bees ... or Not Two Bees?
« Reply #7 on: Sep 26th, 2016, 8:37am » |
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on Aug 24th, 2016, 5:49pm, JiNbOtAk wrote:| (...) If only my teacher had used Grimbal's approach, it would had been far easier to understand (rather than brute-forcing through with the mathematical approach) |
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If your teacher used my approach he wouldn't be a teacher.
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