Physics, asked by kajal365, 10 months ago

A drunkard is walking along a straight road. He takes 5 steps forward and 3 steps backward, followed by 5 steps forward and 3 steps backward and so on. Each step is one meter long and takes one second. There is a pit on the road 11 meter away from the starting point then after how much time drunkard will fall into the pit?

Answers

Answered by narshimhakripa
0

Answer:

he will fall in the pit after 48 sec.

Explanation:

he is taking 5 step forward and 3 steps backward means total after taking 8 steps or in 8 sec ( one sec each step ) he is walking 2 steps ( 5-3) and the pit is 11 steps far so he will need 6 times ( 5 forwad and 3 backward ) and 6*2= 12 so he is taking 2 steps in 8 sec so he will reach to pit in 8*6 = 48 sec

Answered by Anonymous
3

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 = 32 + 5 = 37 s

I hope, this will help

Explanation:

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