Two trains, each having speed 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 60 km apart. On reaching the other train, the bird directly flies back to the first train and so forth. What is the total distance the bird travels before the train collides?

							Ans: Total time interval for the birds motion = time taken for the trains to collide 
 
     ie t = initial seperation / velocity of approach
          = 60/ 2(30)
          = 1 hour

        therfore, distance covered by bird = t.v
                                           = 1.60 = 60 kms
						

							
hiii
when the train is 60 km apart bird flies
time taken by it to travel to other train 
60t + 30t = 60
t = 2/3 hr
when bird reach othe train  distance between two trains = 20 km
time taken by it to travel to first train
60t + 30t = 20
t = 2/9 hr
this process will continue n times
so distance moved by bird = 60 x 2/3 +60 x 2/32 ................................60 x 2/3n
= 60 x 2/3 (1 + 1/3 + 1/32 ......................1/3n )
= 80/3 km    ....................          Ans    
 
						

							
when the separation between two trains is 60km, after which the sum of distance travelled by each train is equal to 60km.
S1 + S2 =60   ( s=ut + 1/2at^2) sice the speec is constat so a=0 )
30t + 30t = 60
t=1hr this is the time taken by the trains to collide after the 60km separation.
so, distance travelled by bird in this time interval is 60x1=60km. ( v of bird=60)