A drunkard walking in a narrow lane takes $5$ steps forward and $3$ steps backward, followed again $5$ steps forward and $3$ steps backward, and so on. Each step is $1 m$ long and requires Is. Determine how long the drunkard takes to fall in a pit $13 m$ away from the start.

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Solution

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.

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