Physics, asked by dharikumarndkm, 1 year ago

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backwards and so on. Each step is 1 metre long and requires 1 second. How long the drunkard takes to fall in a pit 13 metres away from the start

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Answered by Anonymous
23
Hope this will help you
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Answered by Anonymous
10

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Distance covered with 1 step = 1 m

Time taken = 1 s

Time taken to move first 5 m forward = 5 s

Time taken to move 3 m backward = 3 s

Net distance covered = 5 – 3 = 2 m

Net time taken to cover 2 m = 8 s

Drunkard covers 2 m in 8 s.

Drunkard covered 4 m in 16 s.

Drunkard covered 6 m in 24 s.

Drunkard covered 8 m in 32 s.

In the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and falls into the pit.

Net time taken by the drunkard to cover 13 m = 32 + 5 = 37 s

I hope, this will help you

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