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

Answer 1

The x-t graph of the motion of the drunkyard can be easily plotted considering the $5m$ forward and $3m$ backward displacements with the increasing time. From the obtained graph, we can draw a horizontal line through the $13m$ ordinate to intersect the graph. Then drawing a vertical line through the intersection point will give the value of the time required.

Complete answer:
Let the drunkard be at the position $x = 0$ at the time $t = 0$.
As given in the question, the drunkard moves forward by taking $5$ steps forward and $3$ steps backward and each step requires a time of $1s$. Also, it is given that he finally falls in a pit $13m$ away from the start, that is, at $x = 13$. Therefore, his x-t graph is as shown below.

From the above graph, it is clear that the drunkard takes a total time of $25s$ to fall in the pit.
If we combine the forward and the backward motion of the drunkard as one interval, then we can see that the displacement covered in this interval is
$s = \left( {5 - 3} \right)m$
$\Rightarrow s = 2m$ (1)
And the time taken is
$t = \left( {5 + 3} \right)s$
$\Rightarrow t = 8s$ (2)
So in the four intervals, the total displacement covered is
$d = 4s$
Putting (1) in the above equation, we get
$d = 4 \times 2m$
$\Rightarrow d = 8m$
And the time is given by
$T = 4t$
Substituting (2)
$T = 4 \times 8s$
$\Rightarrow T = 32s$
The remaining $5m$ cannot be a part of this cycle, since to cover this displacement, the drunkard only moves forward, not backward. Since each meter is covered in a time of $1s$, so the time required to cover the remaining $5m$ is equal to $5s$. Thus, the total time is given by
$T' = T + 5s$
$\Rightarrow T' = 32s + 5s = 37s$
Hence, the time taken by the drunkard to fall in the pit is equal to $37s$.

Note:
Do not attempt this question by calculating the average velocity of the drunkard. As we can observe in the x-t graph obtained above, the graph between $x = 0$ and $x = 13m$ is not a straight line. So the concept of average velocity is not applicable here.