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sally fields

Guest

problem with dogs

i am having great difficulties in solving the following question. i would really appreciate it if someone with the kindness from their heart and the genius from their brain would help me solve this extremely difficult conundrum.

In a race, each of four greyhounds runs at its own constant speed. they all start at the same point at a circular track; at 30 seconds into the race, while all are still on their first lap, they ahve spread out so they are at four corners of a square. How many seconds into the race will they be when they are next at the four corners of a square?

thanks darlin'
love sally <3

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: problem with dogs

Hi sally fields;

If the first one occurred in 30 seconds the next one is due:

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John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: problem with dogs

Hi, I'll give this one a little try as well:

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Nehushtan

Member

Registered: 2013-03-09

Posts: 957

Re: problem with dogs

For this problem, it is easier to work with angular displacement and velocity rather than the linear equivalents. Suppose the slowest dog travels at angular speed ɷ radians per second. After t = 30 the angular displacements of the of the four dogs, from the slowest to the fastest are:

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If the angular speed of the fastest dog is ρ and that of the second fastest is σ, then

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When the square forms again, after time T say, the fastest dog will have lapped the slowest dog and will be diametrically opposite the second fastest dog. Hence:

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Hence T

So the square formation will occur again one minute after the first, i.e. 90 seconds into the race.

Last edited by Nehushtan (2016-03-06 09:59:31)

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Grantingriver

Member

Registered: 2016-02-01

Posts: 129

Re: problem with dogs

It is much easier than what you have thought of. Since all the greyhounds have constant speeds therefore the difference between the first and the last greyhounds will change at a rate 3/4/30. Now the first possible formation of the square (if it could form) will be when the difference between the first and the last is d₃=1/4 (after the first lap since the difference between the last and the starting point is less then a quarter of a circle) hence: 0.25=3/4/30×t₁-1 ⇒ t₁=1.25/(3/4/30) ⇒ t₁=50 s
But in this case the difference between the first and the second greyhounds will be: d₁=1/4/30×50=5/12 lap which is not a multiple of a quarter of a circle. So let's move to the next chance, this will occur when the difference between the first and the last is 2/4 circle, and the time t₂ to accomplish this will be: 2/4=3/4/30×t₂-1 ⇒ t₂=1.5/(3/4/30) ⇒ t₂=60 s. In this time the differences between the first and the second and the third respectively will be: d₂=1/4/30×60=0.5 lap and d₃= 2/4/30×60=1 lap
Since all the differences between the three greyhounds and the first one are multiples of a quarter circle therefore 60 s is the next time, after the first, in which they form a square again (90 s from the start).

Last edited by Grantingriver (2016-03-06 14:12:41)