When two trains moving with a velocity of 30 km/hr in directions opposite to one another are 60 km apart , a bird starts moving from the top of one train to another with constant velocity of 60km/hr and comes back again. Find the no. of times the bird is able to do so before the two trains collide? (Please solve it)

Hi,

The trains are moving towarsd each other at a speed of (30+30) = 60 kmph.

Distance between them is also 60 km. So the trains will collide in 1 hour.

The bird continuously keeps flying...so in 1 hr the bird will also fly 60 km, because its speed is also 60 kmph

(V=bir v)(v=trn v)(l=sepraton)

1stly vt+Vt=l....then l-2vt=lnew and so on

u will gwt a infinit gp srs

velocity of train 1 = v1 = 30Kmph

velocity of train 2 = v2 = 30Kmph

since train 2 is directed in opposite direction velocity = -30Kmph

relative velocity V of train1 with respect to train 2 = velocity of train 1 - velocity of train 2

= 30 - (-30) = 30+30 = 60Kmph !

so each train is moving with a relative velocity of 60Kmph !

distance = 60Km

time = 60/60 = 1hour

the trains will collide in 1 hour !

bird flies at a speed of 60Kmph so bird covers 60Km in 1 hour !

the trains cover equal distances in equal intervals of time !

the distance covered by each train once the bird reaches another be x then by the time bird flies back it will be 4x because both trains cover x distance ..

train1 ==>|     x     |==>                   <==|       x       |<== train 2

so bird travels 60-4x distance !

so time taken for 1 travel to and fro = 1-(x/15)

no of times bird travels = present distance - (1/x-15) until the two trains collide