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

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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=20−dt -------------(1)

and s=ut+1/2 at2

2=20t−1/2×d×t2 -------------(2)

Putting dt=20 from equation 1 in equation 2,

2=20t−10t

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

Let the magnitude of deceleration be d

Applying equation of motion on the train,

v=u+at

0=20−dt -------------(1)

and s=ut+1/2 at2

2=20t−1/2×d×t2 -------------(2)

Putting dt=20 from equation 1 in equation 2,

2=20t−10t

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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