How Late Did the Man Arrive to Catch the Train?

[PeroK]

Interesting. How?

The easy case, counting the engine as the first carriage (same length):

If the train starts from rest, the time $t_n$ for $n$ carriages to pass satisfies:

$nL = \frac12 a t_n^2$

The time for the n+1 st carriage to pass is then:

$T_{n+1} = t_{n+1} - t_n = \sqrt{\frac{2a}{L}}(\sqrt{n+1} - \sqrt{n})$

Hence you have:

$\frac{T_{n+2}}{T_{n+1}} = \frac{\sqrt{n+ 2} - \sqrt{n+1}}{\sqrt{n+1} - \sqrt{n}}$

That's going to be different for every $n$, so you just need to work out for which $n$ that equation holds.

Now, if there's an engine of length $K$ ...

 

[Charles Link]

The easy case, counting the engine as the first carriage (same length):

If the train starts from rest, the time $t_n$ for $n$ carriages to pass satisfies:

$nL = \frac12 a t_n^2$

The time for the n+1 st carriage to pass is then:

$T_{n+1} = t_{n+1} - t_n = \sqrt{\frac{2a}{L}}(\sqrt{n+1} - \sqrt{n})$

Hence you have:

$\frac{T_{n+2}}{T_{n+1}} = \frac{\sqrt{n+ 2} - \sqrt{n+1}}{\sqrt{n+1} - \sqrt{n}}$

That's going to be different for every $n$, so you just need to work out for which $n$ that equation holds.

Now, if there's an engine of length $K$ ...

@PeroK If you choose an $L$, e.g. $L=30$ m, (and tell us what it is), and choose your own $a$, e.g. $a=1$ m/sec^2, but don't tell us what it is, if you give us $t$ and $t'$ (in seconds) from your formula, (you are free to select $n$ but don't tell us what it is), I believe our calculations for $T$, $a$, and $s=\frac{1}{2}aT^2$ would allow us to readily compute both the value for $a$ as well as the number of cars in the train that had already passed as the passenger arrived. Such a test would basically be showing that our algebra is correct. $\\$ Edit: I tried it using $n=9$ (and 10, and 11), and $a=1$ and $L=30$, and it computed like it should.

 

[haruspex]

some relation between t,t′,t′′

(t'-t-t")³=t'³-t³-t"³?

 

[Philippe Verdy]

@Philippe Verdy We know that. You are making it overly complicated then. We need to assume he arrived just as he observed the edge of one car right in front of him. Otherwise, we have no way of computing anything precisely.

If you assume that, then there's no general solution with the rigid train !

I repeat what we know: we have a unique parabolic arc with vertical passing by three nodes in the cartesian plane, and the horizontal axis of distance is subvided at regular intervals of length L. The measured times are not constrained and can be any real value.

But then the extreme point of the unique ellipse passing by the three nodes is NOT (in the general case) located (on the distance axis of coordinates) at an exact multiple of L because that coordinate is a zero of a quadratic function. We then have a contradiction in the problem or we must seek other reasons:
- the clock is not exact but the error margins allows an imprecision for the determination of the position of the extreme point of the eliipse that includes a region of the 2D plane where it falls between two integer multiples of L.
- if the clock is precise enough and there's no integer multiple L for locating that point of the elliptic arc, the only solution is to admit elasticity. If we suppsoe that the train is rigid (when measured inside the train, then the elasticity is what the traveler observes outside the the train in the station: he must see the effect of relativity.
- now if the three mesures give 3 different locations for the extreme point of the ellipse, the elasticity is no longer an hypothese, it is the only solution and must be observed in the train as well as in the station, independantly of general relativity (whose we can predict the effect precisely).

The important point is that the solution cannot respect the condition that all cars must have the same length L, because it's impossible in the general case! Or the curve of movement is not elliptic, i.e. the acceleration "a" cannot be constant indefinitely !

And we know that acceleration cannot be constant because general relativity cannot support speeds higher than c, so the elliptic movement is certainly wrong (it is only possible as an approximation at low speeds): the two branches of the curve of positions are necessarily decelerating, because the derived curve of speeds is not a straight line, but is a sigmoid whose two branches will necessarily be converging to -c and +c. But, let's ignore the first branch which concerns the train before he arrived in the station there remains only the positive half-sigmoid branch for speeds ! The general relativity applies a negative acceleration on top of the positive acceleration ignited by the engines of the train.

And this is an interesting result, because the traveller with his own precise-enough clock can measure the effect of general relativity, just by observing the passing train (or equivalently by taking place in an elevator falling freely in front of stages of a building, and measuring the time when it passes in front of eah of them; here also we know that it cannot fall freely at constant acceleration, because the acceleration will be zero at the center of Earth and will become negative to continue "falling" to the other side of Earth; the elevator will reach its maximum speed which is about 28000 km/h in the middle of Earth after about 20 minutes and will then decelerate to come to the surface of at the antipode, where it will emerge and will not go above the initial height from where it was initially falling).

We can then observe general relativity on Earth by just looking at trains accelerating in front of us: this just requires a precise enough clock.

And on short distances in a laborary, we can also observe general relativity by measuring the time it takes for a small accelerated graduated wheel to measure interval of times where the regularly spaced graduations passes in front of a sensor: we should see significant differences depending only on the initial angular position of the wheel where it started rotating, and on the precision of the clock.

 

[gneill]

@Jenny Physics : Have you achieved in this thread, to your satisfaction, your goals in solving your mathematical puzzle?

 

[Jenny Physics]

@Jenny Physics : Have you achieved in this thread, to your satisfaction, your goals in solving your mathematical puzzle?

Yes thank you all. I learned quite a bit.

 

[gneill]

Yes thank you all. I learned quite a bit.

Thank you @Jenny Physics . I believe that this thread has served it's original purpose. If others would like to continue the discussion of issues, suppositions, or conundrums related to the premise of the original puzzle, I suggest that that discussion be taken to one of the technical forums. The homework question here has been resolved.

I am closing this thread as it has served its initial purpose.

If anyone feels that there is something substantive and relevant to add to the solution of the original problem as posed, please feel free to PM a mentor to request that your submission be added.