Question

4. A drunkard is walking along a straight road. He takes
5 steps forward and 3 steps backward and so on. Each
step is 1 m long and takes 1 s. There is a pit on the road
11 m away from the starting point. The drunkard will
fall into the pit after
(a) 21s
(b) 29 s
(c) 31 s
(d) 37 s​

Answer 1

Answer:

after walking 6 m it will fall in another 5 m

Answer 2

Answer:

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indiankhiladi

Secondary School Physics 5+3 pts

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

Report by Dharikumarndkm 29.07.2018

Answers

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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

Explanation: