Two trains are headed towards each other on the same straight line, each having a speed of ` 30 km h^(-1)`. A bird that can fly at ` 60 km h^(-1)` flies off one train when they are ` 60 km` apart and leads directly form the other train, On reaching the other train, to flies back to the first train and so on.
(a) How many trips can the bird make from one train to the other train before they meet ? (b) What is the total distance the bird travels ?

Explanation

The velocity of one train relative to the other,
`v_(T T) = (v) - (-v) = 2v = 2 xx 30 = 60 km//hr`
And as the distance between the trains is 60 km, the two trains will crash after `60 km/60 km hr^(-1) = 1hr`.
Now the velocity of bird w.r.t. train towards which it is moving will be `60 + 30 = 90 km//hr`.
so, the time taken by bird for I trip, `t_(1) = (60//90) = 2//3 hr`.
In this time the trains have moved towards each other `(2//3) xx 60 = 40 km`, so the remaining distance `= 60 - 40 = 20 km`.
so, the time taken by bird for II trip, `t_(2) = 20//90 = (2//3^(2)) hr`.
Proceeding in the same way time taken by the bird for `n` the trip, `t_(n) = (2//3^(n)) hr`.
(a) Now if the bird makes `n` trips till the train crashes,
`t_(1) + t_(2) + t_(3) +.....+ t_(n) = 1hr`
i.e., `(2)/(3) + (2)/(3^(2)) +.... + (2)/(3^(n)) = 1 hr`
or `(2)/(3)[1 + (1)/(3) + (1)/(3^(2)) +.....+ (1)/(3^(n - 1))] = 1 hr`
or `(2)/(3) [(1 - (1//3)^(n))/(1 - (1//3))] = 1 hr " or " 1-((1)/(3))^(n) = 1`
or `3^(n) = oo`, i.e.,
So, the bird will make infinite trips.
(b) The distance moved by the bird in I trip `= vt_(1) = 60 xx (2//3) = 40 km` and the distance moved by the bird in II trip `= vt_(2) = 60 xx (2//3^(2)) = (40//3) km`.
So, the total ditance moved by the bird,
`40 [1 + (1)/(3) + (1)/(3^(2))+.....oo] = 40 [(1)/(1 - (1//3))] = 60 km`