Multiple choice

Directions: Answer the question independently. A train of length 220 m is travelling with a constant speed of 90 km/hr. At time t = 0, a bird flies at a constant speed of 108 km/hr from the front end of the train to its rear end and after reaching the rear end of the train immediately flies back and reaches the front end at time t = T seconds. The train advanced by 'y' m during this period. What is the value of 'y'?

  1. 1200

  2. 1000

  3. 1500

  4. 1600
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A Correct answer
Explanation

The relative speed of the bird to the train is 108 - 90 = 18 km/hr (5 m/s) when going to the rear, and 108 + 90 = 198 km/hr (55 m/s) when returning. Time to reach the rear is 220 / 5 = 44 seconds. In 44 seconds, the train moves 90 km/hr * (44/3600) hr = 1.1 km = 1100 m. The bird is now at the rear. To return to the front, the bird must cover the train's length plus the distance the train moved. This is a relative speed problem where the bird catches up to the front. The time taken is 220 / 55 = 4 seconds. Total time T = 44 + 4 = 48 seconds. Distance train moved y = 90 km/hr * (48/3600) hr = 1.2 km = 1200 m.

AI explanation

The relative speed of the bird with respect to the train when flying from the front to the rear is the difference in their speeds: 108 kmph minus 90 kmph, which equals 18 kmph. Converting this to meters per second requires multiplying by 5/18, giving a relative speed of 5 m/s. The time taken to reach the rear end is the length of the train divided by this relative speed: 220 divided by 5, which is 44 seconds. When flying back to the front, their relative speed is the sum of their speeds: 108 plus 90, resulting in 198 kmph, which converts to 55 m/s. The time to reach the front is 220 divided by 55, giving 4 seconds. The total time for the round trip is 48 seconds, during which the train advances at 90 kmph (25 m/s) for a distance of 25 multiplied by 48, which equals 1200 meters.

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