0

Two trains, each having a speed of 30km/h, are headed at each other on the same straight track. A bird that can fly 60km/h flies off the front of one train when they are 60km apart and heads directly for the other train. On reaching the other train, the bird flies directly back to the first train, and so forth. What is the total distance the bird travels before the trains collide?

$D(t) = 60 - 60 t$ is the distance (km) between the two trains after $t$ hours.

$B(t) = {D(t) \over 90} = {2 \over 3} - {2 \over 3} t$ is the time (h) that takes for the bird to fly from one train to the other while the trains are moving, at the moment when $t$ hours has past.

$t_n$ is the time (h) elapsed when the bird has finished the $n$th flight.

We can define $t_n$ recursively as,

$$t_n = t_{n - 1} + B(t_{n - 1}) = {1 \over 3} t_{n - 1} + {2 \over 3}$$

$$t_n - 1 = {1 \over 3} (t_{n - 1} - 1)$$

$$t_0 = 0$$

Let $u_n = t_n - 1$

$$u_n = {1 \over 3} u_{n - 1} = ({1 \over 3})^n u_0$$

$$t_n - 1 = ({1 \over 3})^n (t_0 - 1)$$

$$\therefore t_n = 1 - ({1 \over 3})^n$$

The total distance the bird travels before the trains collide is,

$$\sum_{n = 1}^{\infty} D(t_n) = 60 \sum_{n = 1}^{\infty} ({1 \over 3})^n = 30$$

However, the result should be $60$ because the two trains collide after 1 hour and the bird travels 60km during 1 hour. Which part did I get wrong?

xiver77
  • 373
  • 1
  • 7
  • Fwiw https://math.stackexchange.com/search?q=two+trains – Anne Bauval Mar 19 '23 at 10:24
  • 2
    $B(t)$ is not clearly defined for any $t.$ It should be a sequence, not a function. And why is $B(t_n)=\frac{D(t_n)}{90}?$ – Anne Bauval Mar 19 '23 at 10:35
  • 1
    @AnneBauval 90 is the bird's speed + the train's speed. When t=0, B(0)=2/3, and it takes 2/3 hours at the initial condition for the bird to reach the other train, and so on. I think B(t) can be defined for any t (in hours). – xiver77 Mar 19 '23 at 10:41
  • Please see this answer: https://math.stackexchange.com/a/1001107/1088689 – D S Mar 19 '23 at 11:02
  • 2
    The total distance the bird travels before the trains collide is not $\sum_{n = 1}^{\infty} D(t_n).$ For instance, the distance it travels from $t=0$ to $t=t_1$ is not $D(t_1)=60(1-t_1)=20.$ It is $60B(0)=60\frac23=40.$ – Anne Bauval Mar 19 '23 at 11:04
  • @AnneBauval Thanks, it was an easy fix with your comment. – xiver77 Mar 19 '23 at 11:38

1 Answers1

0

From @AnneBauval's comment I realized that the resulting distance between the two trains does not equal to the actual travel distance of the bird.

$$\sum_{n = 1}^{\infty} D(t_n)$$

should instead be

$$60 \sum_{n = 0}^{\infty} B(t_n) = 40 \sum_{n = 0}^{\infty} ({1 \over 3})^n = 60$$

and the answer is as expected!

xiver77
  • 373
  • 1
  • 7