Question

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again 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.

Answer 1

Hint: We will first find out the net displacement and time taken by the drunkard in each cycle of his movement. Then we will continue to find net displacement until it is short of five steps from the pit as the drunkard will fall before the last cycle completes. Then we can add the time taken in each step to get the total time.

Complete step by step solution:
It has been given that the man takes 5 steps forward and 3 steps backward and thus repeats this cycle. The length of each step is 1m and time taken to cover each step is 1s.
So, in each cycle the drunkard moves 2 steps forward or his net displacement is 2m. And as he takes a total 8 steps (5 forwards + 3 backwards), so the total time taken by him in completing one cycle or covering a displacement of 2m is 8s.
Therefore, average velocity of the drunkard = $dfrac{2}{8}$ = 0.25 m/s.
Let us now find out the net displacement of the drunkard after each cycle which takes 8s to complete,
So, after first cycle, net displacement = $0.25\times 8$= 2m
after second cycle, net displacement = $0.25\times 16$= 4m
after third cycle, net displacement = $0.25\times 24$= 6m
after first cycle, net displacement = $0.25\times 32$= 8m
As, after covering 8m, the drunkard will again move 5 steps or 5m forward and will fall into the pit located at 13m.
So total time taken by the drunkard to reach and fall into the pit = 32 + 5 = 37s.

Note: If we just divide the distance of the pit by the average velocity, then the answer will come as 52 seconds and which will be wrong. So, we must see that when the net displacement is 5m short of the pit distance as when the drunkard will again move 5 steps forward, he will fall into the pit.