Question

A drunkard is walking along a straight road. He takes 5 steps forward and 3 steps backward and so on. Each step is $1m$ long and takes $1s$. There is a pit on the road $11m$ away from the starting point. The drunkard will fall into the pit after:
$\begin{aligned} & \text{A}\text{. }29s \\\ & \text{B}\text{. }21s \\\ & \text{C}\text{. }37s \\\ & \text{D}\text{. }31s \\\ \end{aligned}$

Answer 1

Each step taken by a drunkard requires a time of 1 second. If we see, after taking 5 steps forward, the drunkard takes 2 steps backward; therefore, the last 5 steps of the drunkard can be kept aside and we can calculate the rest of the distance to be covered and the time taken by the drunkard to cover that distance. We can then add both the times, for the initial journey and the last five steps, to get the final answer.

Complete step by step answer:
Given that, drunkard moves 5 steps forward and 3 steps backward in an interval of 8 seconds. After that, he is $2m$ away from the initial position.

We get,
$x=5-3=2m$ in 8 seconds
$x=2m$ is the displacement of the drunkard in 8 seconds
Velocity of the drunkard,
$v=\dfrac{\text{Displacement }}{\text{Time}}$
We have,
$\begin{aligned} & \text{Displacement = }2m \\\ & \text{Time = }8s \\\ \end{aligned}$
Therefore,
$v=\dfrac{2}{8}=\dfrac{1}{4}m{{s}^{-1}}$
Now,
Time taken by drunkard to cover first $6m$ of his journey is given by,
$\text{Time = }\dfrac{\text{Displacement}}{\text{Velocity}}$
Distance to be covered, or displacement, is $6m$
And velocity is $\dfrac{1}{4}m{{s}^{-1}}$
Therefore,
Time taken by drunkard to cover $6m$ distance is,
$t=\dfrac{6}{\dfrac{1}{4}}=6\times 4=24s$
The pit is $11m$ away from the initial position of drunkard. After covering $6m$ distance, the drunkard needs to take 5 more steps and he will fall into the pit.
As given, each step takes 1 second to cover the distance; it means that after travelling $6m$ distance in 24 seconds, the drunkard will take 5 more seconds to fall into the pit.
Total time taken by drunkard is,
$t=24+5=29s$
The drunkard will fall into the pit after $29s$
Hence, the correct option is A.

Note:
Students should not get confused between the distance covered by drunkard and the displacement of the drunkard. Furthermore, the speed of the drunkard should not be confused with his velocity. For falling into a pit, a drunkard needs to cover those 11 meters distance, which is drunkard’s displacement, and not the distance covered by him. Students have to work with velocity term instead of speed term for the drunkard, as velocity is associated with the total displacement, while speed is associated with the total distance covered by a body.