A drunkard walking along a road takes 8 steps forward, 6 steps backward and so on. Each step is 1m long and of 1 sec. How much time will he take to fall in a pit 18m away?
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the correct answer is 126sec
Answer:Brainly.in
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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
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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: