Question

A drunkard walking in a narrow lane takes 8 steps forward and 6 steps backward , followed again by 8 steps forward and 6 steps backward and so on . Each Step is 1 metre long and requires 1second . Determine how long the drunkard takes to fall in a pit 18m away from the start.

Answer 1

Step 1: Given that:

The number of steps taken by the drunkard in forward direction = 8 steps

The number of steps taken by drunkard ion backward direction = 6 steps

Distance covered in each step = 1m

time taken = 1s

Step 2: Calculation of time taken by drunkard to cover a distance of 18m:

Distance covered in 8 steps in forward direction = $8\times 1m=8m$

Thus,

Time taken to move first 8 m forward = $\text{8s}$

Distance covered in 8 steps in backward direction = $6\times 1m=6m$

Time taken to move 6m backward = $6s$

Now the net disp[placement of the drunkard = Distance covered in forward direction - Distance covered in the backward direction

= $8m-6m$

= $2m$
Total time taken to cover the displacement $2m$ = $8s+6s$

= $14s$
That is;

Time taken by drunkard to complete first 2 m displacement = $14sec$

Therefore,

Time taken by drunkard to complete a total displacement of $10m$ = $\frac{14s\times 10m}{2m}$

= $14\times 5$

= $70s$

Now the rest distance of 8m is covered by the drunkard in next 8 steps.

Therefore,

Time taken to complete next 8 m= $8sec$

Thus,

Total time taken to cover 18m =Time taken to complete displacement of 10m + time taken to complete next 8m

= $70sec+8sec$

= $78sec$

Thus,

The drunkard will take $78sec$ to fall in a pit 18m away from the start.