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.

By BYJU'S Exam Prep

Updated on: September 25th, 2023

It is given that

Distance covered in one step = 1 m

Time taken = 1s

So the time is taken to move the first 5 steps forward = 5 s

Time is taken to move 3 steps backward = 3 s

Net distance covered = 5 – 3 = 2m

Net time is taken to cover 2 m = 8 s

Table of content

1. 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.

So we get

Drunkard covers 2 m in 8s

Drunkard covers 4 m in 16s

Drunkard covers 6 m in 24s

Drunkard covers 8 m in 32s

In the next 5s drunkard will cover 5m and a total distance of 13m and fall into a pit

Net time taken by drunkard to cover 13m = 32 + 5 = 37s

x-t graph of the drunkard’s motion is shown as

Therefore, the drunkard takes 37s to fall into a pit 13 m away from the start.

Summary:

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.

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. The drunkard takes 37s to fall in a pit 13 m away from the start.

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