Jumping on a Train: Tips and Advice

[haruspex]

That is what I wanted to confirm, as some of the responses wanted me to give an answer and justify it and I was getting more confused each time I looked at the question and thought that there was a correct option.

But I don't think everyone on the forum thinks that this question and option don't make sense. Btw the answer given by the person who set it is A.

Initially I narrowed it down to A or B but thought A and B are equivalent so to speak so was wondering how to distinguish between them.

The mathematical method is too much for me but it would be interesting which position the formulae give.

I withdraw my previous remarks in this thread.
Just plot the path the man will take and the paths points A to D will take. Remember to keep the speed about the same for the man and the train, except that point A will travel a bit further and point C not quite so far, depending on the radius.

 

[Steve4Physics]

Hi. Thought I’d throw in my tuppence-worth!

I can't read the diagram. Here’s a version of the question which I think is equivalent to what is intended, plus my answer. Maybe it will help.
_____

Question:

A platform (P) is moving at 2m/s north.
P contains an xy grid (initially the y-axis is north).
A man (M),stands at (0,0) on the grid.

At t=0, M jumps up and remains in the air for 1s (lands at t=1s).

At t=0.5s the platform instantaneously rotates 90º clockwise and now moves at 2m/s east (note the y-axis is now east.)

What are M’s coordinates on the xy grid when he lands?
_______

Answer:

At t=0.5s, M is at max. height and is above the point (0,0) (because his velocity relative to P is zero from t=0 to t=0.5s).

Between t=0.5s and t=1s, M has moved a further 2m/s*0.5s = 1m north (now the -x direction) while the platform has moved 1m east.

In the platform’s frame of reference, between t=0.5s and t=1s, M has moved 1m in the -x direction and 1m in the -y direction. So M’s final coordinates are (-1, -1)m.

 

[kuruman]

Yes, as I wrote in post #19 previously.

Can you then comment what is wrong with the simple mathematical model in post #25? It conforms to the parameters of the question as stated in that
The man takes off from point A.
It is possible for the train to complete the 90^o turn.
Regardless of the man's time of flight, there is only one of the listed points where the man lands the closest. This is answer A as OP has revealed.

Furthermore, the model checks out in the limiting cases:
As the time of flight goes to zero, the distance between the landing point and A goes to zero.
When the acceleration is zero, we have the ballistic cart solution, namely that the man lands at the take off point regardless of the time of flight unless, of course, he hits the train wall before landing.

Needless to say, this is not a unique model; it is the simplest one that I could conjure up.

 

[haruspex]

Can you then comment what is wrong with the simple mathematical model in post #25? It conforms to the parameters of the question as stated in that
The man takes off from point A.
It is possible for the train to complete the 90^o turn.
Regardless of the man's time of flight, there is only one of the listed points where the man lands the closest. This is answer A as OP has revealed.

Furthermore, the model checks out in the limiting cases:
As the time of flight goes to zero, the distance between the landing point and A goes to zero.
When the acceleration is zero, we have the ballistic cart solution, namely that the man lands at the take off point regardless of the time of flight unless, of course, he hits the train wall before landing.

Needless to say, this is not a unique model; it is the simplest one that I could conjure up.

I have since withdrawn that post and updated post #19. See also post #31.

Wrt post #25, I don't see how the acceleration can be modeled as constant.

Please note that the description in post #1 is incorrect. See the diagram in post #6. The man starts at point C, which is to the right.
In your post above, you seem to have A as start and finish... I guess you meant takes off from C.

We don't know where the centre of the turn is, neither how far to the right nor how far forward or back from point C, nor where the mass centre of the carriage is (the one point that we can be sure should travel the same arc distance as the man). We are mot told it is 90 degrees, merely that it is sharp, so a radius not too much greater than the separation of the points in the diagram, I guess.

Suppose the man moves distance y in the original direction.
If we assume point B of the train maintains roughly constant speed then it can't go y in the original direction, much less point C. It's less clear, but I cannot find a way for D to get there either. Point A can get there because it arcs at a greater radius, so moves faster.

 

[jbriggs444]

The man takes off from point A.

How can we agree on a solution when we cannot even agree on the problem statement. Post #6 specifies a take-off at point C.

 

[kuruman]

How can we agree on a solution when we cannot even agree on the problem statement. Post #6 specifies a take-off at point C.

OK. He takes off from C. Thank you both @haruspex and @jbriggs444 for pointing that out. That's what I initially thought, but then before I wrote down the equations I thought I'd better recheck the statement of the problem and I looked at #1 which states that he takes off from A. I committed an error thinking that I caught myself in error. In short, a comedy of errors. Back to the drawing board.