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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,056

Two trains travelling with constant speed V towards each other and distance between them is L initially.A bird is flying continueously between them with its speed V'.

1)Find the no. of times bird went to the other train and came back.

2)Find the time after which the trains meet.

I took out the time as L/2V.

Pls tell me about the first part.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,016

I guess we have to assume V' > V or there's no question.

This means the bird can always go between the trains, even when they are close. It looks like, in this idealised situation, the bird has an infinite number of shorter and shorter journeys. I say idealised because we have to assume that it takes no time to turn around and go the other way, and that the gap is never so small that there's no space for the bird.

I hope that's right.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

Yes, if L' (length of the bird) is 0, it is an infinite number of times. If not, then it gets a bit more complicated. All I can say for that case is that after t=(L-L')/(2V) time, L' starts monotonically decreasing.

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,056

yes it is infinity but I wanted to know how to frame equations for these type of questions.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,016

hi Niharika,

Let's say the bird is at train A and about to fly towards B. It's tricky as B is moving too.

One way to simplify the problem is to 'bring B to rest'. Imagine a frame of reference which travels with B at V. That effectively cancels out B's velocity so it is no longer a moving target. But you must apply the same to train A and the bird.

new velocity for the bird = V + V'

new velocity for A = 2V

At the moment the bird starts off, B is stationary at a distance L, so the time to reach B is T1 = L/(V+V')

and in that time A will have travelled 2VT1 so that the distance between the trains is now = L - 2VL/(V+V')

Now reverse the bird. Bring A to rest, so that B is travelling at 2V and the bird again at V+V', but now in the other direction.

You can get an expression for T2, the time for the bird to travel to A, and calculate the new distance between A and B.

You should end up with a geometric series, for which you can calculate the sum to infinity.

The algebra gets pretty nasty which may be why Agnishom's link is to a numeric example.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

Hi Bob

Shouldn't the distance end up being finite? Maybe I've just misunderstood what you said.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,016

hi Stefy,

calculate the sum to infinity.

Did you mean this ?

It means

rather than

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,056

thanks bob.

friendship is tan 90°.

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