Two trains, each having a speed of `30 km//h`, are headed towards each other

Explanation

Since the speed of bird `=2xx`(speed of train), if the train covers a distance `x`, then the distance covered by the bird `=2x`.
After the first trip: Bird is on train 2.

The separation between the trains becomes one-third of the original separation and the bird travels two-third of `60 km`.
Separation between trains `=60/3=20 km`
Distance traveled by the bird `=2/3xx60=40 km`
After the second trip: Bird is on train 1.
Separation between trains `=1/3xx20=20/3 km`
Distance traveled by the bird `=2/3xx20=40/3 km`
After the third trip: Bird is on train 2.
Separation between trains `=1/3xx20/3=20/9 km`
Distance traveled by the bird `=2/3xx20/3=40/9 km`
In this manner, the separation between the trains becomes `20/27 km, 20/81 km`, ..., () m ..., () cm, ...() nm and the bird travels `2/3xx20/9, 2/3xx20/27 ...`
Since we treat all objects as point objects, so the separation between the trains will be zero after infinite trips.
Number of trips `= oo`
Total distance traveled by the bird
`=40+40/3+40/9+20/27+...oo`
`=40(1+1/3+1/3^(2)+1/3^(3)+...oo)`
Geometric progression
`=40(1/(1-1/3))=40xx3/2=60 km`