I simply thought of this trouble the other day.
Assume there is a vital sector of straight line
AB that requires to be seen in all time. A guard dog can see in one instructions before itself and also have to stroll at a constant non-zero rate in all time. (All guard dogs do not require to have the very same rate. ) When it gets to completion of the sector, it has to reverse (at no time ) and also maintain seeing the line.
The amount of guard dogs does it require to assure that the line sector is seen in all time? And also just how (first placements and also rates of the pets )?
Keep in mind :
It's clear that 2 pets are not nearly enough. I opinion that 4 will certainly be adequate and also 3 will certainly not. As an example, the listed below arrangement does not function from 7.5 2nd if
AB's size is 10 meters.
Dog 1 at A walks to the right with speed 1.0 m/s Dog 2 at between A and B walks to the right with speed 1.0 m/s Dog 3 at B walks to the left with speed 1.0 m/s
Or it can be highlighted as :
A ---------------------------------------- B 0.0 sec 1 --> 2 --> <-- 3 2.5 sec 1 --> <-- 32 --> 5.0 sec <-- 31 --> <-- 2 7.5 sec <-- 3 <-- 21 -->
Please give your remedies, tips, or relevant troubles specifically in greater measurements or looser problems (guard dogs can stroll with velocity, and so on )
Three pets is enought I assume.
Allow the size of line sector amount to 1 (with works with from 0 to 1).
First pet : start placement = 0, rate =+1/ 3
Second pet : start placement = 2/ 3, rate =+1/ 3
Third pet : start placement = 2/ 3, rate = - 1/ 3
After 1 2nd the placement comes to be comparable.
I'll make the unimportant solution : 1 pet at factor A, encountering factor B, strolling with a rate of 0. Probably, you need to actually highlight that the pets' rates have to be non - absolutely no ... this is the sort of side instance that mathematics individuals enjoy to manipulate.