Bees and Trains: A distance problem

[putongren]

Homework Statement

Consider two trains moving in opposite directions on the same track. The trains start simultaneously from two towns, Aville and Bville, separated by a distance d. Each train travels toward the other with constant speed v. A bee is initially located in front of the train in Aville. As the train departs Aville, the bee travels with speed u>v along the track towards Bville. When it encounters the second train, it instantaneously reverses direction until it encounters the first train, then it reverses again, etc. The bee continues flying between the two trains until it is crushed between the trains impacting each other.
The purpose of this problem is to compute the total distance flown by the bee until it is crushed.
Assume that the bee is faster than the trains.

Relevant Equations

Distance = Rate * Time

This question is from the MIT Courseware. I’m having difficulty finding the general equation to solve the problem

(1). d = vt + ut
d = (u + v)t
t = d/(u + v)

(2). d = vt + ut
d - vt = ut sub t with d/(u+v)
d - (v*d)/(u+v) = (u*d)/(u+v)

I’m done with the easy part. Deducing the general equation is a lot harder. Please give me a hint on how to approach the task of determining the general equation.

 

[kuruman]

Assume that the initial separation is L.
Can you find an expression for the time of the collision?
Can you find the distance the bee has covered in that time?

Clearly, the answer is a number times L.
I don't know what you mean by "general equation."
d = vt is general enough.

 

[Orodruin]

Hint: This is one of those cases where “doing the infinite sum” is a bad bad idea. You might technically be able to do it that way, but it really helps to stop and think for a bit.

This problem is a real classic.

 

[PeroK]

Time is of the essence!

 

[Frabjous]

Hint: This is one of those cases where “doing the infinite sum” is a bad bad idea.

Depends on who you are.

 

[brotherbobby]

I'd suggest the problem be done both as an infinite (geometric) series as well a simple distance-speed-time problem.
To the OP, for what total time do you think does the bee move to and fro between the trains before it meets its end?

 

[Orodruin]

Depends on who you are.

I am going to go ahead and assume that OP is not Gauss considering that Gauss has been dead for almost 170 years.

 

[Frabjous]

I am going to go ahead and assume that OP is not Gauss considering that Gauss has been dead for almost 170 years.

There’s a famous Von Neumann story. When posed with a variant of this question, he immediately answered with the correct result. When asked about knowing the short cut, he said that he had summed the series.

 

[brotherbobby]

I am going to go ahead and assume that OP is not Gauss considering that Gauss has been dead for almost 170 years.

I am no Gauss myself. But I have done this problem as an infinite geometric series that needed some attention and thinking. I didn't need more than 2 to and fro rounds to see the series beginning to form.

But why do a harder method when a shorter but cleverer one is available? I suppose it's healthy practice for the beginner. Long winded techniques have a bad reputation for being silly. But they could be just as satisfying and necessary if one wants to develop the habit of doing long calculations.

More to the point, a Gauss would not do the problem that way. He'd go for the smarter option.

 

[putongren]

After thinking about the problem along with your helpful input, logically, the problem ends when the trains collide.

d/2 = vt -> t = d/(2*v)

Since "t" is the total time it takes for the problem to end, multiply “t” with u (bee’s velocity) to find the total distance the bee travels. Therefore the distance = (d*u)/(2*v)

Out of curiosity, how would you use infinite summation to solve this problem?

 

[Hill]

After thinking about the problem along with your helpful input, logically, the problem ends when the trains collide.

d/2 = vt -> t = d/(2*v)

Since "t" is the total time it takes for the problem to end, multiply “t” with u (bee’s velocity) to find the total distance the bee travels. Therefore the distance = (d*u)/(2*v)

Out of curiosity, how would you use infinite summation to solve this problem?

You get a geometric series.

 

[PeroK]

Out of curiosity, how would you use infinite summation to solve this problem?

The first leg of the journey takes a time $t_1 = \frac {d}{u+v}$. At which point, the trains are a distance $d_2 = d - 2vt_1$ apart. I'll leave it as an exercise for you to simplify $d_2$ in relation to $d, u$ and $v$.

The second leg takes a time $t_2 = \frac {d_2}{u+v}$. And $d_3 = d_2 - 2vt_2$.

This gives you an infinite series of times $t_n$ to add up. As mentioned above, this will turn out to be a geometric series.

 

[brotherbobby]

Out of curiosity, how would you use infinite summation to solve this problem?

Fix one of the trains for simplicity, say the one on the left and forget it thereafter. Let that train be the origin (O) for all measured distances. The train on the right, say T, is moving towards it with a speed $-2v$.

(1) The bee flies from O to T with a speed $+u$. How long will they take to meet? Where will T be when they have met?
(2) From there on, the bee flies back along the $-x$ direction to O. How long will it take? Where will T be when the bee is back at O?

Far as I remember, you have to repeat (1) and (2) twice and you'd be able to "see" a geometric series forming. The series will be an infinite one which we know how to sum.

 

[Orodruin]

Fix one of the trains for simplicity, say the one on the left and forget it thereafter. Let that train be the origin (O) for all measured distances. The train on the right, say T, is moving towards it with a speed $-2v$.

(1) The bee flies from O to T with a speed $+u$. How long will they take to meet? Where will T be when they have met?
(2) From there on, the bee flies back along the $-x$ direction to O. How long will it take? Where will T be when the bee is back at O?

Far as I remember, you have to repeat (1) and (2) twice and you'd be able to "see" a geometric series forming. The series will be an infinite one which we know how to sum.

This is incorrect. You cannot change to a coordinate system where one train is fixed. In that system distance travelled by the fly will be different because its speed is no longer u in both directions. It is u-v in one direction and u+v in the other.

 

[brotherbobby]

This is incorrect. You cannot change to a coordinate system where one train is fixed. In that system distance travelled by the fly will be different because its speed is no longer u in both directions. It is u-v in one direction and u+v in the other.

Excellent point. Distances are not invariant in galilean relativity, though lengths are.
I did this problem for a man and a dog, going home, the dog running back and forth. The home was fixed to the ground.
I think it's instructive to do the problem the way I described if only to see the mistake.

 

[Hill]

There is yet another way to solve this problem, with linear algebra. It does not involve a geometric series but is perhaps above the "Introductory Physics" level.

 

[jbriggs444]

After one leg, the problem has been reduced to a new separation at the same relative speeds. The recursion is evident and leads to a geometric series.

 

[Hill]

determining the general equation

With the linear algebra, I got the following general equation for the total distance flown by the bee after $n$ legs:
$l_n=d\frac {u} {2v} (1-(\frac {u-v}{u+v})^n)$

 

[putongren]

Using Linear Algebra to solve this problem boggles the mind. Did you use matrices or vectors?

 

[Hill]

Using Linear Algebra to solve this problem boggles the mind. Did you use matrices or vectors?

A matrix, its eigenvalues and eigenvectors.