Question:
$2$ cars travelling at $108,75$ kmph respectively, start from the opposite ends of a $90$ km straight road. Travelling back and forth continuously for $10$ hours, how many times do the $2$ cars meet? [No loss of time at the turns]
My Approach:
Time to meet for the first time = $\left(\frac{90}{108+75}\right)$ hours and for any subsequent meeting, they'll take $2\times\left(\frac{90}{108+75}\right)$ hours.
So, for the $n^{th}$ meeting they 'll take a total of $\frac{(2n-1)90}{183}$ hours which should be less than $10$ and solving:
$$\frac{(2n-1)90}{183}\leq10$$
$$\Rightarrow n=10$$
So, the $2$ cars will meet $10$ times.
Solution:
The faster car covers $1080$ km in $10$ hours which is $12$ times the length of road and in each length, it will meet the other car exactly once – in opposite or the same direction.
So, the $2$ cars will meet $12$ times.
My approach neglects the possibility of the $2$ cars meeting while moving in the same direction at some point of time. But that's how I was taught to solve such problems so why the method won't work?? I was only told to be wary of ratio of $2$ objects not $\geq 2$. Are there other limitations too?
But the solution too could be wrong if both the cars, at some point of time, start in the same direction from one of the either ends of the road in which case the cars won't meet each other in that round, but the next round. How could the solution be sure that, that won't be the case??
Just to confirm what's the actual answer I drew the graph to know that the solution's answer is right and I was wrong:
One can see from the above graph that the $2$ graphs meet a total of $12$ times between $0$ and $600$ mins.
Also going a little further, at $1800$ mins, the 2 cars are at the same point and start moving in the same direction so obviously such a thing is possible.
Best Answer
You have already uncovered what I think are the most important aspects of this problem: first, that the $2d/(v_1+v_2)$ formula omits the meetings where the faster car overtakes the other; second, that the "it will meet the other car exactly once" method overcounts meetings that occur exactly at one end of the road. (It counts the arriving and departing trip along the road as two meetings, but in reality there is only one meeting.)
But I think we extend the graphical method to make the necessary calculations a little more obvious. I will start first by using "road lengths" as the unit of linear measurement, so the $90$ km straight road is $1$ road length long. I will also use hours as the unit of time.
In those units, the graph of the two cars going from one end of the road to the other ($0$ units to $1$ unit, or $1$ unit to $0$ units) as time passes looks like this:
This is adequate to do the analysis, but there's a trick that sometimes gives insight into problems like this, where the trajectories essentially reflect off lines (in this case the lines at $y=0$ and at $y=1$). The trick is to make a mirror image of the trajectories, like this:
So basically we are still watching the cars move between the two ends of the road ($y=0$ and $y=1$), but we are also watching their reflections in a mirror that we have placed along the line $y=1.$
If we also place a mirror along the line $y=0,$ we can see a reflection of the reflection:
And more reflections beyond that, as far as we care to look:
So far this is getting more complicated, not simpler. But notice that we only really need to observe when the cars meet each other in order to count the times they meet, and every meeting of the cars is reflected at exactly the same time in each reflection. We really only need to see the meeting in one reflection.
We can do this by tracing just the path of the faster car in just one of the reflected images for each additional length of the road it traverses, like this:
That cleans up the diagram a lot. We no longer have all those reflections of each time the cars meet, but we still have all the times at which the cars meet.
Now continue this for the full $10$ hours. We end up needing $12$ copies of the graph on the road (the original and $11$ reflections) in order to follow the faster car along a single straight line on the graph.
Now counting the intersections looks somewhat like counting the squares that a diagonal line goes through in a rectangular grid, for example as asked in How many squares is passing the diagonal, How many squares does the diagonal of this rectangle go through? or How many crossings are there? Is there a formula? The only difference really is that the lines of the grid are sloped and not horizontal or vertical.
(Note that we could just as easily--and perhaps more helpfully--have labeled the positions along the vertical axis $d,$ $2d,$ $3d,$ and so forth instead of $1,$ $2,$ and $3.$ I used integers because it was more convenient in the graphing tool.)
With a graph link this it becomes particularly easy to see that the opposite-direction meetings must occur at the interval $2d/(v_1 + v_2),$ because when the red line crosses a downward-sloping black line, the next downward-sloping black line is exactly $2$ road-lengths (a distance $2d$ in the original units) above that intersection on the graph and the two lines converge at a rate equal to the difference in their slopes. Similarly, we can see that the overtaking meetings must occur at the interval $2d/\lvert v_1 - v_2\rvert,$ because (again) whenever the red line crosses an upward-sloping black line, the next upward-sloping line is exactly $2$ road-lengths above it, but in this case the rate of convergence is the difference in the slopes.
The similarity to the grid-diagonal problem should also remind us that the red line might cross the grid exactly at an intersection, and that when it does we have to count only one crossing instead of two (one crossing for each line it crossed there).
Since the cars have speeds $5/6$ and $6/5$ in the units of the graphs above, the number of hours between opposite-direction meetings is $$ \frac{2}{\frac56 + \frac65} = \frac{60}{61} $$ and the number of hours between overtaking meetings is $$ \frac{2}{\frac65 - \frac56} = \frac{60}{11}. $$
Since $11$ and $61$ are relatively prime (in fact, both are prime numbers), the least common multiple of those two times is $60$, so if the red line intersects both directions of the grid at once, it will happen every $60$ hours. But we start with the red line exactly in the center of one of the black parallelograms (halfway between the lines above and the reflections below not shown in the graph), and we are guaranteed that in $60$ hours the red line will once again be exactly the center of of a parallelogram. The entire graph has a $180$-degree rotational symmetry about the midpoint of those two points (we could rotate everything $180$ degrees and the red line would remain where it was and the black lines would trade places with each other). This implies that at $30$ hours, the red line will also be at a point of symmetry in the set of black lines: the center of a parallelogram, the center of a side of a parallelogram, or a crossing of two lines. As it turns out, it is at a crossing of two lines. This tells us that the "it will meet the other car exactly once" method is not only logical invalid, it actually will fail once every $60$ hours, the first time being $30$ hours after the starting conditions.
A couple of things we might do in order to make the grid even easier to use would be to do a shear transformation of the plane that makes the upward-sloped black lines horizontal, followed by a shear that makes the downward-sloped black lines vertical. Then the transformed red line would be passing through a rectangular grid in the usual way and the intersections would be easy to count for any period of time, whether that period allows the faster car to traverse the road an integer number of times or not. But this may be more transformation than we really need, since you already have the tools to analyze the problem thoroughly by finding the time to the next meeting and the time between meetings for opposite-direction meetings, overtaking meetings, and meetings at either end of the road.