What Is the Bird's Velocity Relative to the Platform and Train Passenger?

[jianxu]

Homework Statement

A passenger in a train moving at 30m/s passes a person standing on a station platform at t= t' =0. Twenty seconds after the train passes, the person on the platform determins that a bird flying along the tracks in the same direction as the train is 800m away. Five seconds later, the bird is 850m away. Assume a constant velocity for all involved. Find the velocity of the bird as determined by the observer on the platform and train passenger.

Homework Equations

x' = x-vt ----- v is the difference of the velocity between bird and train, t is the same for passenger or person on platform, x is the location of the bird in the frame of the person standing on the platform so(x2 = 850 and x1 = 800)

v'=(x2'-x1')/(t2'-t1')

The Attempt at a Solution

So what I did is just plugged in the information they gave us so that x1' = 800 - 20m/s(difference between train and bird)*20s and x2' = 850 - 20*25

lastly I found velocity by v'=(x2'-x1')/(t2'-t1') and got -10m/s as my answer.

The negative portion seems logically right since the passenger in the train will see the bird come closer and closer as if the bird's velocity is the opposite. I'm just not sure if everything is mathematically correct and was wondering if someone can do a check for me? Thanks a lot!

 

[mal4mac]

I don't get the same answer.

My powers of visualisation are limited so I have to work these kinds of question out step by step, so here's my working:

t = 0s
------
v = 30m/s velocity of train, throughout
------
------
t = 20s
------------
b = 800m bird's distance from platform
-----
-----

t = 25s
-------
b = 850m bird's distance from platform

Therefore: bird travels 50 metres in 5s, giving bird's velocity relative to observer on platform of 10 m/s

Velocity of bird relative to train follows easily: -20 m/s

Check my working! I always make slips in these kinds of things.

 

[jianxu]

ok I think I understand this. So what we actually want to do is not apply gallilean transformation to the problem. We took the difference of the bird's speed and the train's speed and in this regard, the speed of the train relative to the passenger inside it will appear to be zero. Next either mathematically(x2-x1)/(t2-t1) or logically, the bird will appear to be going the opposite direction with a velocity of -20 m/s. In the mathmatical sense I took the difference between the train and the bird at time t to be the x component and was able to achieve the -20m/s. Is this the correct way to think/solve this problem?

 

[MeJennifer]

The speeds in the above example are not relativistic but since the topic indicates this is a special relativity question one might want to use the relativistic velocity addition formula here.

 

[Doc Al]

I assume, despite the title, that this is meant to be solved using the usual Galilean relativity, not Special Relativity. (I assume that you'll be contrasting this with what is needed at much higher speeds in a later problem.) You can certainly solve this using special relativity as well, but I doubt that's what they want.

x' = x-vt ----- v is the difference of the velocity between bird and train, t is the same for passenger or person on platform, x is the location of the bird in the frame of the person standing on the platform so(x2 = 850 and x1 = 800)

There are two frames here. The platform frame (unprimed) and the train frame (primed).

The equation x' = x - vt, relates position measurements in those two frames. v is the speed of the train with respect to the platform (v = 30m/s).

What equation relates velocity measurements in those two frames? Call the speed of some object (the bird, in this case) by the letter u (to distinguish it from v, the frame speed). Find an equation relating u' and u.

 

[jianxu]

Yes I think I did meant galilean relativity sorry. So to relate u' and u I would say u' = u - v and u' is the bird's velocity in the frame of the passenger in the train which would just be 10m/s - 30m/s which is -20m/s ?

 

[Doc Al]

Yes I think I did meant galilean relativity sorry. So to relate u' and u I would say u' = u - v and u' is the bird's velocity in the frame of the passenger in the train which would just be 10m/s - 30m/s which is -20m/s ?

Exactly right.

 

[jianxu]

thanks for the help =P

 

[mal4mac]

ok I think I understand this. So what we actually want to do is not apply gallilean transformation to the problem.

You might be able to proceed that way. What does the situation look like if you consider the (i) bird at rest (ii) train at rest. It is is useful to get thinking about different objects defining the rest frame at non-relativistic speeds -- what I know as Galilean relativity:

http://en.wikipedia.org/wiki/Galilean_relativity

We took the difference of the bird's speed and the train's speed and in this regard, the speed of the train relative to the passenger inside it will appear to be zero.

Unless he's going for a coffee :-)

Next either mathematically(x2-x1)/(t2-t1) or logically, the bird will appear to be going the opposite direction with a velocity of -20 m/s. In the mathmatical sense I took the difference between the train and the bird at time t to be the x component and was able to achieve the -20m/s. Is this the correct way to think/solve this problem?

Looks like your getting it. If you are uncertain: think about the situation, draw it, write down what you're thinking in your own words., break down difficult steps into smaller ones... Keep doing this until you don't need to ask anyone else if your way is correct -- you'll know it is!

George Polya - "How to solve it" is a great book on general problem solving. One of his best tips is "solve a simpler problem first". So here you might think of just the train and bird. If train is going at v and bird is going at u, how fast is the train going relative to the bird? Easy, yeah? Making up little problems like this can get you clearer on the bigger problems.

P.S. Doc Al's use of frames and primes is probably more standard than my approach. But, hey, whatever gets the problem solved!