Some mathematical problems are beautiful not because of their answer, but because they can be approached in more than one way. The following puzzle is one such example. One method leads us through the fascinating world of infinite geometric series, while another arrives at the same answer through a simple observation. Exploring both approaches reminds us that mathematics is not just about reaching the destination—it is also about appreciating the different paths that lead there.
Two trains are 100 km apart and moving toward each other, each at 50 km/h. A bird starts from the front of one train and flies toward the other at 100 km/h. When it reaches the second train, it instantly turns around and flies back toward the first train. It continues flying back and forth until the trains collide. How far does the bird fly?
Many people try to calculate the distance of each individual trip, creating an infinite series. But this problem can be solved instantly by understanding it thoroughly. We will see the shortcut later. Let us see the straightforward approach of building the series.
Let's work through the infinite-series approach for this problem.
Problem Setup
Distance between trains = 100 km
Each train speed = 50 km/h
Bird speed = 100 km/h
The bird starts from Train A and flies toward Train B.
First trip
The trains are initially 100 km apart.
The bird and Train B are moving toward each other with relative speed
100 + 50 = 150 km/h
So, the time for the first trip is
t1 = 100/150 = 2/3 h
Distance flown by bird:
d1 = 100*(2/3) = 200/3 km
During this time each train moves
50*(2/3) = 100/3 km
Thus, the distance between trains shrinks by
(100/3) + (100/3) = 200/3 km
Remaining distance:
100 - 200/3 = 100/3 km
Second trip
Now the bird flies back toward Train A.
The bird and Train A again approach each other at relative speed
100+50=150
So, the time for the second trip is
t2 = (100/3)/150 = 2/9 h
Distance flown by bird:
d2 = 100*(2/9) = 200/9 km
During this time each train moves
50*(2/9) = 100/9 km
Thus, the distance between trains shrinks by
(100/9) + (100/9) = 200/9 km
Remaining distance:
(100/3) - (200/9) = 100/9 km
Third trip
Similarly,
the time for the third trip is
t3 = (100/9)/150 = 2/27 h
Distance flown by bird:
d3 = 100*(2/27) = 200/27 km
During this time each train moves
50*(2/27) = 100/27 km
Thus, the distance between trains shrinks by
(100/3) + (100/3) = 200/27 km
Remaining distance:
(100/9) - (200/27) = 100/27 km
Pattern
One can notice the pattern here. Distance between the two trains is reducing. So, the time required by the bird to fly from one train to other too is reducing.
The bird's successive distances are
first trip 200/3,
second trip 200/9,
third trip 200/27 and so on.
Total distance travelled by the bird
This is a geometric series in which a term is getting multiplied by 1/3 to get the next term.
We find the sum of this.
d = (200/3) + (200/9) + (200/27) + (200/81) + -------
Multiplying both sides by 1/3, we get
(d/3) = (200/9) + (200/27) + (200/81) + -------
Subtracting both, we get
(2d/3) = 200/3
This gives d 100 km.
Elegant Shortcut
This problem can be solved instantly by understanding it thoroughly.
- Relative speed of the trains = 50 + 50 = 100 km/h.
- Time until collision = 100 km ÷ 100 km/h = 1 hour.
- Bird flies for exactly 1 hour at 100 km/h.
Therefore:
A nice generalization
Suppose:
initial distance = L
each train speed = u
bird speed = v
Then the bird's distances form
d = vL/(u+v) + (vL/(u+v)) * ((v-u)/(u+v)) + (vL/(u+v)) * ((v-u)/(u+v)^2) + -----
whose sum simplifies to
d = vL/2u
Also, since the initial distance between trains is L and the trains approach each other with relative velocity of 2u, collision time is (L/(2u)). The bird keeps flying for this time at a speed of v. Hence the distance covered by the bird is
d = vL/2u
which is the elegant shortcut solution.
The infinite-series method is longer, but it beautifully shows how an infinite number of trips can add up to a finite distance.
Well Done
Now think about following situation:
Two cyclists are riding towards each other on a straight road at speed of 15 km/h. They are initially 30 km apart. A dog runs back and forth between them at 25 km/h until they meet. What distance dog runs?
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