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...
There's another difficulty in this problem: it's clear that even if we restrict the conditions to use only 2 times measurements, the solution for T is a zero of a quadratic equation.
Now let's project the origin of the parabol at time -T along the x axis: we get an x0 position which is NOT...
It's extra info that allows checking the solution, or verifying that there was no elastic elongation or that all cars of the train have the same length L (if the train is inelastic)
So now assume we don't know L: it's the same as saying that the x vertical axis has no known unit. We can use any...
There's a way to represent the problem geometrically: if the acceleration of the train is constant, the graph of the position x of the start of the train according to time forms an vertical parabolic curve. The 3 measurements are made at regular intervals (of length L) along the vertical x axis...
I don't know where the formulas used by gneill to compute successive speeds come from "magically".
The problem speaks about a train where the passenger clearly knows the number of cars (he was arriving in the station where he expected to see the first car before it left the station: how does he...
@haruspex, you're wrong on BOTH your claims.
The problem specifies "v_l" this limits the number of cars because "v_l" is already the speed reached at the station where the train started to go.
This implies a fixed number of cars which can "tend to infinity", so the time per carriage cannot tend...
"constant acceleration of the train" ? So why I was conter-argued that we needed 3 time measurements? The experiment exposed in the problem is already testing the elastic model if it really needs the 3 time measurement variables (in which case there's NO constant acceleration of the **whole**...
The comments was about the contest that claimed there was a need of 3 time variables in the problem.
I argue that only 2 are needed but only under some strict inelastic conditions that actually never exist in any real train.
At least 3 time variables are needed for the minimum elastic problem...
Also for trains that transport very heavy weights, the cars cannot be safely built to be completely rigid. They would not resist to elongation forces and would crack. Instead, the structure is built with an elastic floor supported by non-rigid, non-straight horitontal bars which are slightly...
You can check by yourself: the substitution works on both formulas to give a solution to your problem, that does depends only on two time measurements (not three). The solution is necessarily the same if there's no elastic elongation of the train. And you've noted yourself that T=v_l/a:
We've...
Note: we must also assume that there's no elastic deformation of cars (this elastic deformation exists for non relativist speeds, but for solid trains it should may be neglected... that's not really true in actual trains where elastic deformation really occurs at each attachment between each...
If acceleration "a" is contant, and each each car has length "L", then it's just enough to use t an t': the difference (t'-t) is proportional to the acceleration "a" and to "L" (and does not even depend on the initial speed of train when it starts going from the station)
We can also say (t''-t'...