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Question

Two trains each having a speed of 30 km/h are headed at each other on the same straight track. A bird that can fly at 60 km/h flies off from one train when they are 60 km apart and heads directly for the other train. On reaching the other train, it flies directly back to the first and so forth. How many trips (n) can the bird make from one train to the other before the trains crash. Find the total distance that the bird travels assuming bird to be a point object.

A
n=,60 km
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B
n=,50 km
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C
n=60,60 km
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D
n=60,50 km
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Solution

The correct option is A n=,60 km
Relative speed of one train w.r.t. another train is vrel=30+30=60 km/hr
Relative distance covered by one train w.r.t. another train is drel=60 km
Time taken for the train to meet
t=6060=1 hour
Relative velocity between bird and train on which bird is approaching is (60 km/h+30 km/h=90 km/h)
For the first trip time taken by bird
t1=6090=23 h
Separation between the trains after first trip of bird
s1=dvrelt1=6060×23=20 km
For the second trip, time taken
t2=2090=232 h
Separation between the trains after second trip of bird is
s2=2060×232=203 km
We can see that for third trip time
t3=233 h
For nth trip tn=23n h
Hence total time is going in geometric series
T=2(13+132+133+.........+13n)
=23(113n)113=113n
But we know that available total time is 1 hour
1=113n
Hence, n= (bird will do infinite number of trips before collision)
Total distance travelled by the bird
vt=60×1=60 km

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