1. Tardigrade
  2. Question
  3. Physics
  4. 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. The time the drunkard takes to fall in a pit 13 m away is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 $1\,s$. The time the drunkard takes to fall in a pit $13\, m$ away is

Motion in a Straight Line

Solution:

The drunkard advances $(+5-3) m =2\, m$ in $8\, s$. Hence $8\, m$ in $32\, s$. Then in the next $5\, s$, he would move $5 m$ more or reach $13\, m$ or the pit. Hence he reaches the pit in $37\, s$.