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.