3
$\begingroup$

Few days ago, I was asked the following question:

There are $2$ cities. city $A$ and city $B$ with distance $d$=600km
There are $2$ trains with speed of $vt$ = 100km/h.
There is $1$ fly with speed of $vf$ = 300km/h.

The question:
Train 1 goes from city A to city B , while Train 2 goes in the opposite direction.(both start at the same time). the fly starts from train 1 and go to train 2 , than back to train 1 , than back to train 2...etc...

What is the distance that the fly passes till the trains cross each other ?

The answer is simple - it takes 3 hours for the train to cross -> fly was running 900km($300 \times 3$).

I was wondering on the "hard" way of solving this problem by actually calculating the distance the fly pass on each stage. It appears to be the sum of a (finite?) series of the distances.

Could you help me solve this problem the "hard way" ?

$\endgroup$
  • 1
    $\begingroup$ Correction: One hour = 3600 seconds. So 3 hours = ? $\endgroup$ – Namaste Aug 12 '14 at 15:33
  • 1
    $\begingroup$ There is a nice discussion of this well-known problem to be found here. $\endgroup$ – Stefan Mesken Aug 12 '14 at 15:34
  • 1
    $\begingroup$ See "Related" column on the right. $\endgroup$ – user147263 Aug 12 '14 at 15:35
  • 2
    $\begingroup$ Those are some seriously fast trains. $\endgroup$ – Fly by Night Aug 12 '14 at 15:36
  • $\begingroup$ @FlybyNight So fast that I almost want to ask in which frame of reference the trains start simultaneously. $\endgroup$ – David H Aug 12 '14 at 15:43
3
$\begingroup$

If the trains are distance $d$ apart and the fly starts on one train then it takes $d/(vt+vf)=d/400$ hours for the fly to reach the other train, and at that time the trains are $(2 vt) d/400 = d/2$ km apart, the fly having flown $(3 d/4)$ km. So after each trip from train to train the trains close by half the distance, so the fly travels $3 d/4+3d/8+3d/16+\cdots$, or $3d/4 \sum_{n=0}^\infty 1/2^n$. Now, the series $\sum_{n=0}^\infty 1/2^n$ is a geometric series of the form $\sum_{n=0}^\infty \alpha^n$ which if $|\alpha|<1$ equals $1/(1-\alpha)$, or in our case 2. So in our case, as $d=600$ the fly travels $2 * 3 * 600 / 4 = 900$km.

The story of von Neumann saying that he summed the series, always struck me as a completely reasonable approach. If you stare at geometric series all day, it is easy to recognize this problem as a series problem, and it is not much more difficult than using the "trick". In a certain sense, the "trick" is implicit in the formula.

$\endgroup$
  • 2
    $\begingroup$ The way I've always heard this anecdote, it was von Neumann who summed the series. I have heard many Norbert Wiener stories, but this wasn't among them. $\endgroup$ – David K Aug 13 '14 at 1:39
  • $\begingroup$ @david You're probably right. $\endgroup$ – deinst Aug 13 '14 at 1:49
1
$\begingroup$

For the $n^\text{th}$ leg of the journey of the fly:

  • let $t_n$ be the time taken
  • at the start of this leg:

    • the total distance already travelled by both trains is $\quad 2v_t\sum_{i=1}^{n-1}t_i$
    • the distance between the fly and the opposite train is $\quad d-2v_t\sum_{i=1}^{n-1}t_i$

Using Speed $\times$ Time = Distance, we have

$$\begin{align} (v_t+v_f)t_n&=d-2v_t\sum_{i=1}^{n-1}t_i\\ 400t_n&=600-200\sum_{i=1}^{n-1}t_i\\ 2t_n&=3-\sum_{i=1}^{n-1}t_i\\ \end{align}$$

Subtracting consecutive terms gives $$\begin{align}2(t_n-t_{n-1})&=-t_{n-1}\\ t_n&=\frac 12 t_{n-1}\end{align} $$ which is a geometric series.

Hence $$t_n=\left(\frac 1{2^{n-1}} \right)t_1=\frac 32 \left(\frac 1{2^{n-1}} \right)=\frac 3{2^n}$$ and distance covered by the fly on the $n^{\text{th}}$ leg is $$s_n=300t_n=\frac {900}{2^n}$$

Total distance travelled by fly, $$\begin{align}S&=s_1+s_2+...+s_n+...\\ &=450+225+...+\frac {900}{2^n}+...\\ &=900\sum_{i=1}^{n}\frac 1{2^i} \\ &=900\end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.