Question

Two trains, each having a speed of 30 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 67 km apart and heads directly for the other train. On reaching the other train it flies directly back to the first train, and so forth. (We have no idea why a bird would behave in this way.) What is the total distance the bird travels before the trains collide?

Answer 1

Now when the bird took off from a train initially the trains were at a distance of 67km Now when the bird moves the train also approaches it So, if we see from frame of reference of the second train the bird has to move with a speed of (60 + 30)km/s relative to the train and travel a distance of 67km. Thus time taken by it will be t = 67/90 hrs.And after than the bird will reverse its position and move towards the other train the same way. In that time the trains had travelled a distance of 2t.30km/hr = 60t = 44.67km. So, in that time the distance between them will change to 67 - 60t = 22.33km . And now again the bird will move the same way and will take time t₂ = (67 - 60t)/90 = 0.248hr and thus the distance between the trains is changed to 22.33km - 60t₂ = 7.44km. So now if you can see clearly we can write an expression of the time elapsed by the bird everytime it changes its direction which can be given as,

67/90 = t

t₂ = t/3

t₃ = t₂/3.....

This means that the time elapsed will be a Geometric series;

of the form t, t/3,t/9,t/27.....

Now this will go for infintely many times thus the sum of the infinite series can be given as,

Total time elapsed by the bird in moving till the train crashed, T_Total = t/(1-1/3) = 3t/2 = 1.5 x 67/90 = 1.11667 hrs

Now the total distance travelled by the bird will be D = 60.T_total = 67km