Motion puzzle: Bird flying between a train and a platform

[kuruman]

This ma
If it takes the train an infinite amount of time to reach the station, sure. I didn't realise that case was under discussion.

Not here. We are given that the braking acceleration is constant which sets the braking time at $T=\dfrac{2L}{v_0}=40~\text{h}.$

 

[erobz]

Not here. We are given that the braking acceleration is constant which sets the braking time at $T=\dfrac{2L}{v_0}=40~\text{h}.$

Which using the idea, puts an upper bound of 144000 collisions with this train given 1 second pauses at the platform and a bird that defies all laws of physics in a "huge" way!

 

[bob012345]

This ma
If it takes the train an infinite amount of time to reach the station, sure. I didn't realise that case was under discussion.

I only mentioned it for completion.

 

[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.

I looked for the maximum distance, D, such that the bird only touches the train once. I.e. the train stops by the bird just as the bird is about to set out for its second journey. I used $v$ for the constant speed of the bird, $a$ for the positive deceleration of train, and $b$ for the time the bird pauses.

The answer, sadly but not unexpectedly, didn't simplify:
$$D = \frac{4v^2}{a} + \frac{ab^2}{2} -6bv + 2\bigg (v\sqrt {\frac 2 a} -b\sqrt{\frac a 2}\bigg )\sqrt{\frac{2v^2}{a} -4bv}$$

 

[bob012345]

I looked for the maximum distance, D, such that the bird only touches the train once. I.e. the train stops by the bird just as the bird is about to set out for its second journey. I used $v$ for the constant speed of the bird, $a$ for the positive deceleration of train, and $b$ for the time the bird pauses.

The answer, sadly but not unexpectedly, didn't simplify:
$$D = \frac{4v^2}{a} + \frac{ab^2}{2} -6bv + 2\bigg (v\sqrt {\frac 2 a} -b\sqrt{\frac a 2}\bigg )\sqrt{\frac{2v^2}{a} -4bv}$$

I think acceleration $a$ is not independent of $D$. I mean it can’t be independently set. There is still the relationship that the acceleration is set by the train initial velocity and $D$.

 

[PeroK]

I think acceleration $a$ is not independent of $D$. I mean it can’t be independently set. There is still the relationship that the acceleration is set by the train initial velocity and $D$.

D is a function of a, v and b.

 

[bob012345]

D is a function of a, v and b.

Isn’t acceleration of the train always $a=\frac{v_t}{2D}$? $v_t$ is the train initial velocity.

 

[PeroK]

Isn’t acceleration of the train always $a=\frac{v_t}{2D}$? $v_t$ is the train initial velocity.

The initial speed of the train is variable in this calculation.

 

[gyorgiy]

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!

No, the one second delay before each round trip makes the number finite

 

[gyorgiy]

This does not appear to be a normal "homework question". It looks like the result of an attempt to "add just a litle difficulty" to a question. However, in this case it seems to me that this "little extra difficuly" transforms the question from one which has a closed-form solution to one which appears to require a sequence of such solutions. (The train's deceleration doesn't really affect this).
I may be wrong, but I personally wouldn't expect this to have a closed-form solution, so I wouldn't tackle it in this way. (I would get serious and tryto find one if I was told it existed - by someone I trusted).
My inclination would be to find a closed from solution for each round-trip, and paste this into something that can perform iterations. (Either a simple program or a spread-sheet would do the job).
BTW, the values chosen are weird: Using brakes to slow down uniformly over a 200-km distance - at this speed, normal losses will stop it within a km or so (on level tracks). We get: average train speed of 5-km/hr, journey time 40 -hours, deceleration 0.25-km/hr/hr etc

 

[gyorgiy]

Right, it bounces perfectly and instantly.

That's a very different (and much simpler) problem than the one posed here. The bird flies for 40 hours, fo it flies 4,000-km. And it makes an infinite number of ever-shorter round trips.

 

[gyorgiy]

Isn’t acceleration of the train always $a=\frac{v_t}{2D}$? $v_t$ is the train initial velocity.

The time it takes is D/(V/2), so the deceleration is V/(D/(V/2), or Vinitial^2/(2D)

 

[gyorgiy]

I looked for the maximum distance, D, such that the bird only touches the train once. I.e. the train stops by the bird just as the bird is about to set out for its second journey. I used $v$ for the constant speed of the bird, $a$ for the positive deceleration of train, and $b$ for the time the bird pauses.

The answer, sadly but not unexpectedly, didn't simplify:
$$D = \frac{4v^2}{a} + \frac{ab^2}{2} -6bv + 2\bigg (v\sqrt {\frac 2 a} -b\sqrt{\frac a 2}\bigg )\sqrt{\frac{2v^2}{a} -4bv}$$

I suspect that you are overcomplicating?
Working backwards, I take it that touching once is the touch when the train is at the platform? Using the numbers backwards we have:
Train starts from platform: Dtrain=0 at t=0, so (D in km, t in hours)
Dtrain = a*t^2/2, where t is the time in hours, and a=0.25m/hr/hr
Bird starts from platform after one second, so (first trip) Dbird = 100*(t-1/3600)
Bird reaches train for the second time when a*t2^2/2 = 100*(t2-1/3600), or
t2^2 - 200*t2/a + 1/18/a = 0
t = 100/a +/-sqrt((100/a^2 - 1/18/a) hours = 1.000000347 seconds
(Note: this is indistinguishable at this precision from the first order approximation = in one second the train moves just 1/4/3600^2/2 = 9.645E-9 = km, which takes the bird 9.645E-9/100*3600 seconds = 0.3472-microseconds)

 

[PeroK]

I suspect that you are overcomplicating?
Working backwards ...

Is a good idea, but I get the same expression. It's not one that I can iterate.

 

[gyorgiy]

Is a good idea, but I get the same expression. It's not one that I can iterate.

If you wish to iterate, you need to define a cycle that starts from the same place each time - for example the bird arriving back at the platform.
In this case, t'2 = 2*t2 - 1
Then each iterative step is finding the value of t(n+1) in:
a*t(n+1)^2/2 = 100*(t(n+1)-t'n)
and setting
t'(n+1) = 2*t'(n+1) - 1

 

[PeroK]

If you wish to iterate, you need to define a cycle that starts from the same place each time - for example the bird arriving back at the platform.
In this case, t'2 = 2*t2 - 1
Then each iterative step is finding the value of t(n+1) in:
a*t(n+1)^2/2 = 100*(t(n+1)-t'n)
and setting
t'(n+1) = 2*t'(n+1) - 1

Easier said than done.

 

[gyorgiy]

Easier said than done.

It's easy enough to turn this into an explicit formula for each iteration.
And easy enough to write a programme or set up a spreadsheet to do the iterations.

I doubt that there is a closed-form formula for train distance versus number of trips (other than one that merely embeds the iterations), but I could be mistaken

 

[PeroK]

It's easy enough to turn this into an explicit formula for each iteration.

Let's see it then!

 

[erobz]

It's easy enough to turn this into an explicit formula for each iteration.
And easy enough to write a programme or set up a spreadsheet to do the iterations.

I doubt that there is a closed-form formula for train distance versus number of trips (other than one that merely embeds the iterations), but I could be mistaken

Please see LaTeX Guide to format your mathematics for readability.

 

[bob012345]

I looked for the maximum distance, D, such that the bird only touches the train once. I.e. the train stops by the bird just as the bird is about to set out for its second journey. I used $v$ for the constant speed of the bird, $a$ for the positive deceleration of train, and $b$ for the time the bird pauses.

The answer, sadly but not unexpectedly, didn't simplify:
$$D = \frac{4v^2}{a} + \frac{ab^2}{2} -6bv + 2\bigg (v\sqrt {\frac 2 a} -b\sqrt{\frac a 2}\bigg )\sqrt{\frac{2v^2}{a} -4bv}$$

Letting $r=\frac{\Large {v_b}}{\Large{v_t}}$, the ratio of the velocity of the bird to the train, I get

$$D=\frac{1}{2}\left( \frac{\large{v_t}b}{-1 -2r +2 \sqrt{r^2+2r}}\right)$$

This is simply setting $2 t_1 + b = T$ where $T$ is the time for the whole trip for the train and using $T= \frac{2D}{\Large{v_t}}$ and $t_1=T(r+1 -\sqrt{r^2 +2r})$.

 

[PeroK]

The original idea was to calculate the total number of touches. A better problem is to have the train leave the station with constant acceleration $a$, the bird wait for some time $b$ before setting off at speed $v$, touching the train, then flying back again at the same speed. Then waiting another time $b$ and repeating.

For any given parameters $a, b, v$ we can do a numerical iteration to count how many times the bird touches the train.

Analytically, the solution depends on the dimensionless parameter $\dfrac v {ab}$.

We can find a condition on this parameter for the bird to touch the train at least once. Trying to iterate analytically beyond that looks difficult or impossible.

 

[bob012345]

The original idea was to calculate the total number of touches. A better problem is to have the train leave the station with constant acceleration $a$, the bird wait for some time $b$ before setting off at speed $v$, touching the train, then flying back again at the same speed. Then waiting another time $b$ and repeating.

For any given parameters $a, b, v$ we can do a numerical iteration to count how many times the bird touches the train.

Analytically, the solution depends on the dimensionless parameter $\dfrac v {ab}$.

We can find a condition on this parameter for the bird to touch the train at least once. Trying to iterate analytically beyond that looks difficult or impossible.

The equation for the time of the first meeting of the bird and train in this scenario is



$$T_1=\frac{v}{a}\left(1-\sqrt{1-\frac{2ab}{v}}\right)$$


If $\large\frac{2ab}{v}>1$ the bird does not touch the train. If it is equal to one, it touches exactly once. If it is less than one it touches at least once. Generalizing the formula for the times the bird touches the train as



$$T_i=\frac{v}{a}\left(1-\sqrt{1-\frac{2a\alpha_i}{v}}\right)$$ where $T_i$ is the $i^{th}$ time it touches. Here is the progression of $\alpha_i$ in terms of the times it touches.



$$\alpha_1=b$$

$$\alpha_2=2T_1$$

$$\alpha_3=2(T_2-T_1)+b$$

$$\alpha_4=2(T_3-T_2+T_1)$$

$$\alpha_5=2(T_4-T_3+T_2-T_1)+b$$

$$\alpha_6=2(T_5-T_4+T_3-T_2+T_1)$$

$$\alpha_7=2(T_6-T_5+T_4-T_3+T_2-T_1)+b$$ ect…



At each progression, one can test if $\large\frac{2a\alpha_i}{v}>1$ and if it is then there is no $i^{th}$ meeting.

 

[bob012345]

Here is a graph of the above for values $v=10$, $a=1$, $b=1$;

 

[PeroK]

I took another look at the related problem. An object starts with constant acceleration $a$. After some time, $b$, another object sets off in pursuit at constant speed $v$. If it reaches the first object, it returns to the starting point at the same speed, waits another time $b$ and repeats.

Can we find an expression for the minimum value of $\mu \equiv \frac{v}{ab}$ such that the object catches the first object $n$ times?

In general, if the first object has a head start of $T_n$ and the second object catches it after a further time $t_n$ then we need:
$$\frac 1 2 a(T_n + t_n)^2 = vt_n$$This gives the equation:
$$t_n = \frac v a - T_n - \sqrt{\big (\frac v a \big )^2 -2\big(\frac v a \big)T_n}$$Or
$$\frac{t_n} b = \mu - \frac{T_n}{b} - \sqrt{\mu^2-2\mu \big (\frac{T_n}b \big)}$$This has a solution iff:$$\mu \ge \frac{2T_n}b$$We have $T_1 = b$, hence (for at least one catch) we need:
$$\mu \ge 2$$In which case:
$$\frac{t_1} b = \mu - 1 - \sqrt{\mu^2-2\mu}$$Next, we have $T_2 = 2b + 2t_1$, hence (for at least two catches) we need:
$$\mu \ge 4\mu - 4\sqrt{\mu^2 - 2\mu}$$Hence$$\mu \ge \frac{32}{7}$$In which case:
$$\frac{t_2}{b} = -\mu + 2\sqrt{\mu^2 - 2\mu} - \sqrt{-3\mu^2 + 4\mu\sqrt{\mu^2-2\mu}}$$In the case that $\mu = \frac{32}{7}$, we have:
$$t_1 = \frac b 7, \ t_2 = \frac{16}{7}b$$Next, we have $T_3 = 3b + 2t_1 + 2t_2$, hence (for at least three catches) we need:
$$\mu \ge 2\frac {T_3}{b} = 6 + 4\frac{t_1}{b} + 4\frac{t_2}{b}$$This leads (hopefully this is correct) to the equation:
$$959\mu^4 -7992\mu^3 + 3048\mu^2 - 224\mu - 16 \ge 0$$This doesn't have a rational solution, but implies that approximately $\mu \ge 7.94$.

As expected, it seems impossible to get a closed form for the nth iteration.

 

[bob012345]

I put this amended problem in Desmos to play with.



One can set the parameters such as $v$ of the bird and slide the index $i$ to see which iterations touch the train or one can set the iteration number for the desired number of touches and slide the parameter to get the minimum value.