1-D Kinematics: Trains and a bird

[opus]

Homework Statement

Two trains, each having a speed of 30km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train, the bird flies directly back to the first train, and so forth. What is the total distance the bird travels before the trains collide?

Homework Equations

All constant acceleration equations:

$v = v_0 +at$
$x-x_0 = v_0t + \frac{1}{2}at^2$
$v^2 = v_0^2 + 2a(x-x_0)$
$x-x_0 = vt - \frac{1}{2}at^2$
$x-x_0 = \frac{1}{2}(v_0+v)t$

The Attempt at a Solution

Haven't made any progress. There seem to be a lot of moving parts so I just started off trying to find when the trains collide which would be when their x position is equal. I can't seem to pin down what values I can use in what equations to get this. Any hints? I've attached an image to show my drawing.

 

[opus]

What I'm thinking is to maybe get something that represents the distance between the trains at any given time t, and use that general expression as a substitute in the equation for the bird.

 

[haruspex]

find when the trains collide

Definitely the right approach. But your attachment is unintelligible. You need to define your variables and explain the reasoning behind the equations.
After time t, how far has each train gone? What isthe remaining distance between them?

By the way, this is not kinematics:
https://en.m.wikipedia.org/wiki/Kinematics

 

[opus]

Thanks for the reply. As for the equations, I'm not sure which to apply. The acceleration is irrelevant in the problem, so anything with acceleration I do not want. This is why I chose $x-x_0=\frac{1}{2}(v_0+v)t$
This hasn't gotten me anywhere because I neither know t nor x final. So my idea now is to get a rate at which they're approaching one another. Naturally, I want to say that they're approaching each other at 30 + 30 = 60 km/h. But I don't have sound reasoning for this because if I add the velocities I get (30 km/h) + (-30km/h) = 0 km/h which is nonsense.

 

[opus]

Another approach I've tried is to take the equation I listed, apply it to each train, and set them equal to each other to get a time at when the positions are equal. I got t=2 which again makes no sense.

Position of train 1 where I set $x_0=-30km$ and final velocity equal to 0.
$$x-(-30)=\frac{1}{2}(30+0)t$$
$$x=15t-30$$

Position of train 2 where I set $x_0=30km$ and final velocity equal to 0.
$$x-(30)=\frac{1}{2}(-30+0)t$$
$$x=-15t+30$$

Setting them equal to each other and solving for t gives t=2.

 

[Delta2]

I believe the hint here is to express the distance that the bird covers say $x(t)$ as a function of the distance that one train covers $y(t)$ in the same time.

I believe you can prove that the bird travels double the distance that one train covers (i can't reveal the whole reasoning but the main reason is that because its speed is double). So it ll be $x(t)=2y(t)$.The time of collision is $t_0=1hour$ (why??). Each train has cover distance $y(t_0)=30km$.

 

[opus]

I believe the hint here is to express the distance that the bird covers say $x(t)$ as a function of the distance that one train covers $y(t)$ in the same time.

I believe you can prove that the bird travels double the distance that one train covers (i can't reveal the whole reasoning but the main reason is that because its speed is double). So it ll be $x(t)=2y(t)$.The time of collision is $t_0=1hour$ (why??). Each train has cover distance $y(t_0)=30km$.

Do you mean something along the lines of treating both trains as a single train which has a rate equal to the combined rate of the two separate trains, and then see how far the bird would have got when the "single" train has covered 60 km?

 

[Delta2]

Do you mean something along the lines of treating both trains as a single train which has a rate equal to the combined rate of the two separate trains, and then see how far the bird would have got when the "single" train has covered 60 km?

No, but I think what you saying its equivalent to what I was thinking. So yes but you got to prove that this approach is equivalent.

 

[opus]

if I add the velocities I get (30 km/h) + (-30km/h) = 0 km/h which is nonsense.

What about this? Knowing that the velocities are opposite and equal in magnitude, I don't feel justified in just saying that the combined rate at which they're approaching each other is 30 + 30 = 60 km/h.

 

[Delta2]

Btw what's the answer key for this problem, if my approach is correct the answer would be 60km. Is this the answer key?

 

[opus]

By the way I'm talking about, if I just add the speeds, I can get:

$x-x_0=\frac{1}{2}(v_0+v)t$
$60km=\frac{1}{2}(60+0)t$
$t=2$ hours

30 km/h for 2 hours is 60 km/h which is the correct answer, but I feel like I just threw stuff at the wall until it stuck.

 

[opus]

Btw what's the answer key for this problem, if my approach is correct the answer would be 60km. Is this the answer key?

Yes that's correct it's 60 km/h according to the key.

 

[Delta2]

Now that I think again this can be solved easily by just thinking at the time t it takes for the two trains to collide (as @haruspex noted too). So how much time it takes for the two trains to collide (I already said 1 hour but why?) and how much distance the bird covers in this time?

 

[opus]

Now that I think again this can be solved easily by just thinking at the time t it takes for the two trains to collide (as @haruspex noted too). So how much time it takes for the two trains to collide (I already said 1 hour but why?) and how much distance the bird covers in this time?

Post #5 is where I attempted this but got 2 hours which is way too long. Twice the time it looks like.

 

[Delta2]

Post #5 is where I attempted this but got 2 hours which is way too long. Twice the time it looks like.

Yes you do a mistake there, you treat the trains like they are moving with constant deceleration such that their velocity is zero exactly the time the collision happens. But this is not what the problem says. The problem says that the two trains are moving with constant velocity towards each other from the beginning till the moment of collision.

 

[Ganit]

Just concentrate on bird , it is moving with constant speed(60 km/h) ; to find the distance covered by any object moving with constant speed : you only need to find time . So find time before ... BANG

 

[opus]

Ohhhh. Ok

Yes you do a mistake there, you treat the trains like they are moving with constant deceleration such that their velocity is zero exactly the time the collision happens. But this is not what the problem says. The problem says that the two trains are moving with constant velocity towards each other from the beginning till the moment of collision.

Ahhhh I see. I think I was looking into this way too much. Like that the bird's distance would get smaller and smaller as the trains got closer and that had to be accounted for.

 

[PeroK]

What about this? Knowing that the velocities are opposite and equal in magnitude, I don't feel justified in just saying that the combined rate at which they're approaching each other is 30 + 30 = 60 km/h.

You should be able to justify this. Just draw a diagram, showing two trains $60km$ apart, both traveling at $30km/h$ towards each other. Then show their positions an hour later.

. So my idea now is to get a rate at which they're approaching one another. Naturally, I want to say that they're approaching each other at 30 + 30 = 60 km/h. But I don't have sound reasoning for this because if I add the velocities I get (30 km/h) + (-30km/h) = 0 km/h which is nonsense.

You must always think about and decide whether to add or subtract. If two objects are moving towards each other you add their speeds to get the relative speed with which they are moving towards each other.

Ohhhh. Ok

Ahhhh I see. I think I was looking into this way too much. Like that the bird's distance would get smaller and smaller as the trains got closer and that had to be accounted for.

You could solve this problem by adding up lots of little journeys for the bird, back and forward. You'd get an infinite series of some description, which you sum. That's not actually a bad exercise if you want a harder problem.

But, sometimes the solution to problem like this is more about logic that physics.

1) You are asked for the total distance, so that is about speed, not velocity.

2) In this problem, what is the speed of the bird as a function of time?

 

[jbriggs444]

Ahhhh I see. I think I was looking into this way too much. Like that the bird's distance would get smaller and smaller as the trains got closer and that had to be accounted for.

You could attack the problem by evaluating a series of trips from one train to the other and summing the resulting geometric series.

There is an amusing story/urban legend that this problem was posed to John Von Neumann who solved it quickly. When asked how he had solved the problem so rapidly, he responded, "by infinite series, of course".

http://thesciencepundit.blogspot.com/2006/07/john-von-neumann-and-mathematicians.html
http://mathworld.wolfram.com/TwoTrainsPuzzle.html
http://mathforum.org/dr.math/faq/faq.fly.trains.html

 

[SammyS]

Post #5 is where I attempted this but got 2 hours which is way too long. Twice the time it looks like.

Think about it.

What distance has each train traveled at the time they collide ?

 

[opus]

Thanks for the replies everyone. It has come down to:
Train 1 position as a function of time: $T_1=-30+30t$
Train 2 position as a function of time: $T_2=30-30t$

Setting them equal, they meet at 1 hour. Thus, the bird flies one one hour at a speed of 60 km/h. So this bird has flown 60 km.
I am going to come back to this once I learn series to do it the other way.