Physics - Kinematics Question with Solution | TestHub
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 required 1 second to cover. How long the drunkard takes to fall in a pit 13 m away from the start?
Answer:
Solution:
To reach 13 m, the drunkard effectively moves m in seconds.
After 5 such cycles, he covers m in seconds.
From 10 m, he takes 3 more steps forward (3 m) to reach 13 m. This takes 3 seconds.
Total time = seconds.
However, the pit is at 13 m. Once he reaches 13 m, he falls. He does not take the backward steps from 10 m if he reaches 13 m in the forward steps.
So, after 4 cycles, he covers m in seconds.
From 8 m, he takes 5 steps forward. These 5 steps cover 5 m, reaching 13 m.
These 5 steps take 5 seconds.
Total time = seconds.