Question

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 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.

Answer 1

it forms kind of sets.when person moves 5 steps forward and 3 steps backward,his displacement is 2 m.this goes and for each 5 steps forward and 3 steps backward,he covers 2 m.

for a pit 13m away,he has to perform these sets =13/2=6.5

at twefth step he will be 1 m away from pit.when he will start his 5 step forward procedure,after covering only 1 step he will fall.

time taken for a set=8 stepsx1 sec=8 sec.

so for 6 sets,time will be=8x6=48sec.

also when we add the other 1 step,total time to fall in pit will be 49 sec.

Answer 2

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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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