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				   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)

7 years ago


Answers : (3)



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 in 1 hr the bird will also fly 60 km, because its speed is also 60 kmph

7 years ago


(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


6 years ago

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

6 years ago

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