Problem 1 on Motion in a Straight Line

Description:

Problem

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.

Solution

Following is the x − t graph of drunkard’s motion.

Part 2 – Analytical Solution

Given in the problem,

• Every 5 s he moves 5 m forward and the next 3 s he moves 3 m backward.
• To fall in the pit, he has to travel 13 m.

Note − He can never fall in the pit during the backward movement because while moving back he is simply occupying the positions he already crossed during forward motion.

Therefore,

Let the total number of forward steps be ′x′.

Hence, total number of backward steps, ′(x − 1)′. [See NOTE]

Therefore,

5x - 3(x - 1) = 13

2x = 10

x = 5

Hence, he took 5 steps forward and 4 steps backward.

Total Time = Total Length × (1s) = 5 × 5 + 4 × 3 = 37 s

Part 2 – Graphical Solution

Continue drawing the graph till x = 13 m, is crossed.

We see that this happens at t = 37 s.

Hence, that is the time taken for him to fall in the pit.

(Notice how the “NOTE” in the last section can be visualized using the graph below. The result can only lie on the longer line showing the increase in displacement.)