Two trains are headed towaads each other on the same straight rrach, each vaving 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 firectly form the otjer train, On reaching the other train, tg flies back to the first train and so on.
(a) How many trips can the bird make from one trin to the other train befre they meet ? (b) What is the total distance the bird travels ?

Explanation

(a) Relative velocit of train (A) relative to train (B)
` = 30 + 30 = 60 km h^(-1)`
Distansce between train ` A and B=60 km`
Time taken by the two trains to meet
` T = (60)/(60) =1 h`
Relative velocity of bird w.r.t. train (B)
` ` 60 + 30 -90 km h^(-1)`
Time taken for first trip, t_1 = ( 60 km )/(90 km h^(-1))` = 2/3 h`
` Now, separation betwwen the two trains just after first trip of bird ` =60 - 60 xx 2//3 =20 km`
Time taken for the second trip
` t_2 + (20)/(90) = 2/9 h = 2/(3^2) h`
Similarly taken for the nth trip
` t_n = 2/ 3^n h`
and time taken for the nth trip
` t-n = 2/3^n h`
and time taken for the nth trip
` t_n = 2/3^n h`
Total time of flight of bird will be
` T= t_1 = t_ 2 + t_3 + .........+ t_n`
`= 1/2 + 2 /3^2 + 2/3^3 + ...........+ 2/3^n`
` = 2/3 [1 + 1/3 + 1/3^2 + ......... + 1/(3^n -1)]`
` 2/3 [ (1- (1//3)^n)/( 1- (1//3 ))] =1 1- (1/3)^n`
[Formula for sum of G.P. series is ,
` S= (a (1 -r^n)/(1 -r) , Here, a = 1 , r = 1//3 ]`
But ` T =1 h, so 1=1 - (1/3)^n or 1/3^n or 1/3^n =0 or n= prop`
(b) Total distance travelled by the bird
` = constant velocity xx time = 60 xx 1= 60 60 km`.