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two persons A and B starts from the opposite ends of a $90~\mathrm{km}$ straight track and run to and fro between two ends. Speed of A is $30 ~\mathrm{m}/s$ and B is $\frac{125}{6} ~\mathrm{m/s}$. They continue their motion for 10 hours. How many times do they pass each other?

I have done in the following way : The speeds of two persons are $108~\mathrm{km/h}$ and $75~\mathrm{km/h}$. The first person covers $1080~\mathrm{km}$ in 10 hours and thus he made 12 rounds. Thus the person will cross another person 12 times in any one of the direction.

Is this answer accurate? or is there any possibility of their not meeting each other in these 12 rounds for being faster?

Jonas Meyer
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user6595106
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    Graphing A's and B's position on a position vs time graph might be a fruitful approach. – turkeyhundt Jul 20 '16 at 14:13
  • @Bye_World I think your answer was correct and recommend reposting it. Running in opposite directions on a circular track they would meet again each time they traveled a combined $90$ km, but on the straight track they have to travel a combined $180$ km. – David K Jul 20 '16 at 14:54

2 Answers2

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Hint:

Here you can see a graph (constructed reducing by a factor $10$ all the numbers in OP).enter image description here

It is the plot of the functions: $$ f(t)=9\cdot \frac{2}{\pi}\left| \arcsin\left(\sin \left(10.8\cdot \frac{\pi}{18}t \right) \right) \right| $$ that represents the motion of $A$ (black), and $$ g(t)=9\left(1- \frac{2}{\pi}\left| \arcsin\left(\sin \left(7.5\cdot \frac{\pi}{18}t \right) \right) \right|\right) $$ that represents the motion of $B$ (red).

The graph shows that $A$ and $B $ meet $12$ times in $10$ hours. The rationale is that $B$ is always in a point of the track and $A$ is faster than $B$, so each time that $A$ completes a track, it meets $B$ just one time, but $B$ can encounter $A$ two times in the time that he takes to complete a track.

Emilio Novati
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You are correct. Each pass that runner $A$ makes she must pass by runner $B$ somewhere. And once runner $A$ passes $B$, there's no way for $A$ to pass $B$ again before $A$ turns around because $B$ is the slower runner. Thus you only need to know how many times $A$ runs each direction which is $12$.

There is a caveat: what happens if they ever both meet at one of the ends? If they met at one of the ends, then that would lower the number of times $A$ passes $B$ by one each time they meet at one end of the track. This is because $A$ is essentially meeting $B$ in both her trip up to the endpoint and away from the endpoint. So let's figure out if they ever meet at one end or the other during the $10\textrm{ hr}$.

Clearly $A$ is at one of the two ends at $\frac 56, \frac{10}6, \dots, 10\textrm{ hr}$. Likewise $B$ is at one of the ends at $\frac 65, \frac{12}5, \dots, \frac{48}{5}\textrm{ hr}$. So, to get the set of points where they might cross at an end$^\dagger$, we can set $$\frac 56k = \frac 65s \\ \implies k=\frac{36}{25}s$$ Because $36$ and $25$ are relatively prime (they share no common prime factors), the smallest positive value of $s$ such that $k$ is an integer is $25$, which does not occur during the timespan of $10\textrm{ hr}$.

So you were right. They pass each other $12$ times. Good job.

$^\dagger$: This method also includes times when they arrive at opposite ends at the same time, but that won't matter in this case.

  • No, that's if they're running opposite directions on a round track. They're running back and forth on a straight track this time. This doesn't take into account the fact that they can pass one another while running the same direction. – Arthur Jul 20 '16 at 14:25
  • @Arthur Fixed.$ $ –  Jul 20 '16 at 15:14
  • can you please tell me a value of speed B considering this problem for which A will pass B again before A turns around? I just want to know such an example case @Bye_World – user6595106 Jul 20 '16 at 19:23
  • @user6595106 The faster runner will never meet the other twice before turning around. The slower will be able to meet the other twice before turning around. In fact it happens twice in this example -- just look at Emilio's graph. –  Jul 20 '16 at 19:30