A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1m long and requires 1s. Determine how long the drunkard takes to fall in a pit 13m away from the start.
Answers
Answer:
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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 =
✔️ Answer ✔️
Your answer:-
He moves 5 steps forward and 3 steps backward in 8sec after that he is 2m away from an initial position
x= 5-3=2m in 8 sec
after 24sec he will 6m away from an initial position
He moves further 5m forward then he falls in a pit which is 11m away from the initial position.
total time T=24+5=29sec.
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