Motion puzzle: Bird flying between a train and a platform

[anthraxiom]

Homework Statement

Suppose there is a train moving towards a platform at 10 km/h. It applies brakes for constant deacceleration such that it stops at the platform. The platform is 200 km away. A bird sits on the platform which flies at 100 km/h. It flies towards the train, touches it and flies back. Then it waits 1 second before repeating the process. How many times does the bird touch the train?

Thanks in advance

Relevant Equations

equations of motion

I thought of splitting the distance travelled by train in cycles where 1 cycle was when the bird starts from platform, touches train and goes back. Since its speed is uniform I thought it would take equal times for it to get to train and to get back to platform.

Now, distance travelled by train in time taken for bird to touch it is, 1/2 aT^2 + ut where u is original speed. Afterwards, the bird goes back and distance travelled by train in this time is: 1/2 at^2 plus vt where v is velocity when bird touched train. v=aT+u.
adding, we get distance travelled by train in 1 cycle is:1/2a(t^2+T^2)+aT +2u
after this i am stuck

 

[haruspex]

Suppose the bird leaves the platform at time $t_0=0$, then $t_1$, etc.
If the train starts with speed $v_0$ at distance $s_0$ from the platform, what is its speed at $t_n$ and where is it?
Can you find a relationship between $t_n$ and $t_{n+1}$?

 

[anthraxiom]

Suppose the bird leaves the platform at time $t_0=0$, then $t_1$, etc.
If the train starts with speed $v_0$ at distance $s_0$ from the platform, what is its speed at $t_n$ and where is it?
Can you find a relationship between $t_n$ and $t_{n+1}$?

i found the relation but is something absurdly untidy which doesnt just simplify out to a clean result. With this i see that with multiple replacement of vn with the precding value the total ratio may be gotten and divided by the total time. Now i had gotten the same thing before but instead of comparing time, i had compared the distance travelled in 1 cycle which is also as complex. my doubt now is how am i supposed to condense the sum of all the terms into 1 concrete formula. I tried this with displacement but i simply couldnt due to the complexity and abandoned that line of reasoning. Please guide me on how to convert the formula of sums into a concrete formula.[/SUB]

 

[PeroK]

You could work backwards. How close does the train have to be for the bird to touch it precisely once? Then generate an expression for the range of distances for precisely $n$ touches.

Just an idea.

 

[pbuk]

Are you sure the numbers in the question are correct? 10 km/h fir the initial speed of the train?

 

[anthraxiom]

Are you sure the numbers in the question are correct? 10 km/h fir the initial speed of the train?

actually i just made up the numbers. i tried to post it without numbers in the physics sub topic one. I dont think the numbers should really matter as i actually wanted to find the result for arbitrary velocities and distances.

 

[haruspex]

i found the relation but is something absurdly untidy which doesnt just simplify out to a clean result. With this i see that with multiple replacement of vn with the precding value the total ratio may be gotten and divided by the total time. Now i had gotten the same thing before but instead of comparing time, i had compared the distance travelled in 1 cycle which is also as complex. my doubt now is how am i supposed to condense the sum of all the terms into 1 concrete formula. I tried this with displacement but i simply couldnt due to the complexity and abandoned that line of reasoning. Please guide me on how to convert the formula of sums into a concrete formula.[/SUB]

Please post the relationship you found and the work leading to it.

 

[anthraxiom]

Please post the relationship you found and the work leading to it.

Let S_n be the distance travelled in the n^th cycle (with the definition of 1 cycle given in the text).

2aS_n=v_n+1² - v_n^2.

S_n=v_n+1²/2a

now we need to find v_n+1 in terms of v_n
v_n+1=v_n+2at

where t is time taken for bird to touch train from platform.
now it is a matter of plugging the expression of t into the equation which i can not do due to formatting issues. To find t, we just need to equate equation for bird and train and apply quadriatic formula)
now, we are left with S₀=Σs_n
we will then plug in the values for sn and then there is where im stuck as i cant figure out how to isolate n in the equation and solve for it.

 

[haruspex]

Let S_n be the distance travelled in the n^th cycle (with the definition of 1 cycle given in the text).

2aS_n=v_n+1² - v_n^2.

So the initial train speed is $v_1$.

S_n=v_n+1²/2a

A $\Sigma$ missing?

now we need to find v_n+1 in terms of v_n
v_n+1=v_n+2at

where t is time taken for bird to touch train from platform.

What about the pause at the end of each cycle?

now it is a matter of plugging the expression of t

What expression can you write for t (or rather, $t_n$)?
And $S_n$ as a function of that and the velocities?

 

[anthraxiom]

What about the pause at the end of each cycle?

oh right i forgot about that, thanks

 

[erobz]

I don't follow how you are attempting to approach this.

This appears to be a challenging problem. I don't have a full solution, but here is how I'm approaching it.

$v_B$ is the speed of the bird
$V_o$ is the initial speed of the train
$a$ is the acceleration of the train

The cycle is this. Let $t_c$ be the time of the first collision:

$$v_B t_c = s_o - V_o t_c + \frac{1}{2}at_c^2 \tag{1}$$

$$ \implies t_c= \cdots $$

So to find the time duration( resetting clock at 0 ) of the next time of collision you input into the same equation (1) the updated values for initial train position $s_o \to s_1$ and initial speed $V_o \to V_1$.

$$ \implies s_1 = s_o -V_o ( 2 t_c + 1 ) + \frac{1}{2}a(2t_c+1)^2 $$
$$ \implies V_1 = V_o - a ( 2 t_c + 1 ) $$

And now repeat...

Then as a constraint the summation of twice the calculated times + 1 has to be less than/equal to the total time the train travels to travel 200 km (less 1 second -the train could arrive while the bird is sitting on the platform).

 

[PeroK]

Where's John von Neumann when you need him?

 

[erobz]

Where's John von Neumann when you need him?

I hope I'm not missing something obvious, but it seems horrific.

This is not like the "how far does the bird travel question, if it stays moving all the time at $v_b$".
...It feels related, but oh so far away...

Maybe an April Fools Joke?

 

[erobz]

My spreadsheet tells me 1875 collisions for the parameters you gave.

 

[anthraxiom]

I hope I'm not missing something obvious, but it seems horrific.

This is not like the "how far does the bird travel question, if it stays moving all the time at $v_b$".
...It feels related, but oh so far away...

Maybe an April Fools Joke?

I actually read that question first before i thought about this one

 

[anthraxiom]

My spreadsheet tells me 1875 collisions for the parameters you gave.

View attachment 359343

I gave the parameters so that i couldn't just brute force through the question by calculating each one but thanks for the graph :D

 

[bob012345]

My spreadsheet tells me 1875 collisions for the parameters you gave.

View attachment 359343

How do you get a cumulative time greater than the total time the train is moving which is 20 hours or 72,000 seconds?

 

[erobz]

How do you get a cumulative time greater than the total time the train is moving which is 20 hours or 72,000 seconds?

I get a total time of 144,000 seconds

The cumulative time in the graph are the partial summations of $2t+1$. Where $t$ is the time it takes for the birds to reach the train once it's left the platform each cycle. The summation matches my solution for the total time the train travels within a second(144,001 s). I've attached the excel file.

 

[bob012345]

I get a total time of 144,000 seconds

View attachment 359375

The cumulative time in the graph are the partial summations of $2t+1$. Where $t$ is the time it takes for the birds to reach the train once it's left the platform each cycle. The summation matches my solution for the total time the train travels within a second.

Sorry, I forgot the train was slightly decelerating!

 

[kuruman]

I get a total time of 144,000 seconds

View attachment 359375

The cumulative time in the graph are the partial summations of $2t+1$. Where $t$ is the time it takes for the birds to reach the train once it's left the platform each cycle. The summation matches my solution for the total time the train travels within a second(144,001 s). I've attached the excel file.

It's probably easier to use the average velocity: $$S=v_{\text{avg.}}~T\implies T=\frac{S}{v_{\text{avg.}}}=\frac{200~\text{km}}{5~\text{km/h}}=40 ~\text{h}.$$

 

[erobz]

It's probably easier to use the average velocity: $$S=v_{\text{avg.}}~T\implies T=\frac{S}{v_{\text{avg.}}}=\frac{200~\text{km}}{5~\text{km/h}}=40 ~\text{h}.$$

Probably?

 

[Lnewqban]

Have you guys considered that, when flying to meet the train, the bird needs to slowdown to a stop in midair, then start accelerating back in the same direction the train is moving, then contact the train at the velocity it has at that instant and keep flying back?

 

[erobz]

Have you guys considered that, when flying to meet the train, the bird needs to slowdown to a stop in midair, then start accelerating back in the same direction the train is moving, then contact the train at the velocity it has at that instant and keep flying back?

I didn’t consider any actual physics(at that level of precision). Just a brain teaser type solution with kinematics.

 

[kuruman]

Have you guys considered that, when flying to meet the train, the bird needs to slowdown to a stop in midair, then start accelerating back in the same direction the train is moving, then contact the train at the velocity it has at that instant and keep flying back?

Yes, the situation is equivalent to the bird flying past the starting point and then taking time $\Delta t=1~\text{s}$ to reverse direction and be back at the starting point. Similarly, another $\Delta t$ reversal time can be added at the train end to make the question more realistic.

 

[erobz]

Yes, the situation is equivalent to the bird flying past the starting point and then taking time $\Delta t=1~\text{s}$ to reverse direction and be back at the starting point. Similarly, another $\Delta t$ reversal time can be added at the train end to make the question more realistic.

You lose that rather nice time symmetry for the return trip if you do.

 

[pbuk]

to make the question more realistic.

Realistic? 40 hours for the train to slow down for a platform?

This is not a "brain teaser", just a pointless spreadsheet exercise - there is clearly no closed form solution.

 

[erobz]

Realistic? 40 hours for the train to slow down for a platform?

This is not a "brain teaser", just a pointless spreadsheet exercise - there is clearly no closed form solution.

"Pointless" is a relative term.

 

[pbuk]

"Pointless" is a relative term.

Yes: relative to puzzles which require some degree of thought to come up with an efficient method, or which have a surprising solution, this one is IMHO pointless.

 

[erobz]

Well, just because I'm not bright enough to come up with something efficient doesn't mean it doesn't exist, or you should give up trying. The function looks like it could be in the family of $f(n) = e^{-n} n!$. Can it be proven to not have a closed form solution, or am I just not pulling the correct levers?

Anyhow, I think most of everything is pointless...so this is nothing new.

 

[kuruman]

Realistic? 40 hours for the train to slow down for a platform?

This is not a "brain teaser", just a pointless spreadsheet exercise - there is clearly no closed form solution.

Well, I said "more realistic" as compared with reversing direction instantly. Also, no realistic bird flies for 40 hours at 100 km/h with one-second breaks between velocity reversals.

 

[haruspex]

@anthraxiom , what is the source of this question? Is it something you or a colleague made up?

 

[anthraxiom]

Well, I said "more realistic" as compared with reversing direction instantly. Also, no realistic bird flies for 40 hours at 100 km/h with one-second breaks between velocity reversals.

i really didn't want to find the answer per se i just wanted to see if an expression could be found which is the cause for the bizarre numbers. I really thought a solution could be attained pehaps.

 

[anthraxiom]

@anthraxiom , what is the source of this question? Is it something you or a colleague made up?

yes my friend did make it up on the fly

 

[haruspex]

yes my friend did make it up on the fly

Then we should leave it to your friend to astonish the world with an analytic solution.

 

[MatinSAR]

Then we should leave it to your friend to astonish the world with an analytic solution.

Shouldn't the sequence of touches form a convergent infinite series?

Edit: I mistakenly thought this was like the classic train-and-bird problem with constant speeds. Apologies for the off-topic question!

 

[erobz]

What you should try to do - is take a step back and try to solve the "easier problem". Let the train travel at constant speed, and remove the rest pause of the bird. See if you are able to find a closed form solution for the number of collisions.

 

[Lnewqban]

You lose that rather nice time symmetry for the return trip if you do.

The flight is still symmetrical.
It is only that the bird touches the surface of the train some time after it stops in front of it and starts flying back.
If its stop is delayed, a painfull collision will follow.

I see it as equivalent to the the advancement of the spark jump in any gasoline internal combustion engine.

 

[erobz]

Shouldn't the sequence of touches form a convergent infinite series?

I agree, but the challenge for the accelerating train and pausing bird ,how far do you have to compute until the pattern becomes clear is the problem.

 

[MatinSAR]

I agree, but the challenge for the accelerating train and pausing bird ,how far do you have to compute until the pattern becomes clear is the problem.

I mistakenly thought this was like the classic train-and-bird problem with constant speeds. With acceleration, it definitely becomes more complicated.

 

[erobz]

The flight is still symmetrical.

The bird will be accelerating during the pause on the train, its position will change. It will have a shorter return trip to close out the cycle.

 

[erobz]

I mistakenly thought this was like the classic train-and-bird problem with constant speeds. With acceleration, it definitely becomes more complicated.

The classic problem is "asking how far the bird travels in that case" which is turns out to be trivial if you stop and consider it. Finding the convergent series for the number of collisions seems (at least - I haven't solved it) to be much more work even in that problem (unless I haven't had an "aha" moment yet).

 

[WWGD]

Strange how this question is swinging back and forth aimlessly, just like the bird.

 

[erobz]

Strange how this question is swinging back and forth aimlessly, just like the bird.

I don't understand why you are apparently so frustrated. Is there something wrong with demonstrating the challenges of the problem by asking the OP to solve the "easier problem". Also, this is also how we make progress and learn stuff...by solving an easier problem. Are you really going to try and complain to have this thread shut down? What else is going on that is so pressing we can't be bothered to teach/learn?

 

[WWGD]

I don't understand why you are apparently so frustrated. Is there something wrong with demonstrating the challenges of the problem by asking the OP to solve the "easier problem". Also, this is also how we make progress and learn stuff...by solving an easier problem. Are you really going to try and complain to have this thread shut down? What else is going on that is so pressing we can't be bothered to teach?

Just a joke, guy. I haven't nor will I report anything.Edit: I am, though, ignoring you from now on. Feel free to reciprocate.

 

[erobz]

Just a joke, guy.

No its not, I'm attacked relentlessly on here by the "gang" for this kind of exploration. What you are doing is signaling that this thread needs to be shut down because I'm participating in it and it has exceeded 44 posts. Do me a favor and report it...so I can be banned. I've been asking for it.

 

[Lnewqban]

The bird will be accelerating during the pause on the train, its position will change. It will have a shorter return trip to close out the cycle.

Sorry, erobz, could you clarify the idea?

Not that it changes anything important in your calculations, but it still seems to me that the round flight, from the point of view of the bird, will take the same time between both moments when its velocity becomes zero (resting point and stopping-coming back point).
The following round flight will take a different time.

Where the train is at during those round flights, seems irrelevant to that fact (time forward = time backward), unless the bird overshoots and collides with the machine.

Of course, if the point of contact bird-train is used as point of reference, the first leg of the flight should take longer (more distance) than the second one.

 

[erobz]

Sorry, erobz, could you clarify the idea?

Not that it changes anything important in your calculations, but it still seems to me that the round flight, from the point of view of the bird, will take the same time between both moments when its velocity becomes zero (resting point and stopping-coming back point).
The following round flight will take a different time.

Where the train is at during those round flights, seems irrelevant to that fact (time forward = time backward), unless the bird overshoots and collides with the machine.

Of course, if the point of contact bird-train is used as point of reference, the first leg of the flight should take longer (more distance) than the second one.

In the case where the bird leaves the platform, intercepts the train, instantaneously flips and comes back to the platform the elapsed time to this point is $T = t_1+t_1 = 2 t_1$

I thought that it was suggested-to fudge in some acceleration effects of the bird meeting the train- that the bird will take off from the platform, intercept the train, rest for 1 second on the train and return to the platform. Here the elapsed time from leaving the platform to returning to it is not $T \neq 2t_1$. It is $T = t_1 + t_2$ where $t_2 < t_1$.

So the symmetry I'm talking about losing is that the round trip for the bird between the platform and the platform is no longer $2 t_1$ , I exploited that symmetry in the equations to calculate where the train is when the bird gets back to the platform. If its messy now, its going to be worse.

 

[bob012345]

In the case of constant velocity and instantaneous acceleration to speed of the bird, and the bird being a mathematical point, isn’t the number of contacts going to be infinite even when the total distance flown and total time flown is finite? That’s because no matter how short the remaining distance, the bird is always faster than the train. Maybe it’s Zeno’s bird?

 

[erobz]

In the case of constant velocity and instantaneous acceleration to speed of the bird, and the bird being a mathematical point, isn’t the number of contacts going to be infinite even when the total distance flown and total time flown is finite? That’s because no matter how short the remaining distance, the bird is always faster than the train. Maybe it’s Zeno’s bird?

Do you mean when the bird takes no pause at the platform?

 

[bob012345]

Do you mean when the bird takes no pause at the platform?

Right, it bounces perfectly and instantly.