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Question

A train is approaching a platform with a speed of 20 km/hr. A bird is sitting on a pole at the platform. When train is 2 km away from the pole, brakes are applied so that the train decelerates uniformly, simultaneously the bird also flies towards the train with a velocity 60 km/hr. It touches the nearest point on the train and flies back and back again and so on. The total distance travelled by the bird before train stops is

A
30 km
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B
15 km
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C
12 km
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D
10 km
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Solution

The correct option is C 12 km
Let the magnitude of deceleration be d
Applying equation of motion on the train,
v=u+at
0=20dt -------------(1)
and s=ut+1/2 at2
2=20t1/2×d×t2 -------------(2)
Putting dt=20 from equation 1 in equation 2,
2=20t10t
t=15 hr

Hence the bird flies with a speed of 60 km/hr for a time of 15 hr
Hence distance traveled by the bird = 60×15=12 km

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