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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

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```										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
```
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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