Verified Solution • 2025–26 Session
moderate
Q2.4: NCERT Physics - Part 1 MOTION IN A STRAIGHT LINE 2.4 Solution: A drunkard walking in a narrow lane takes 5 steps forward and 3 …
⚠️ Common Exam Traps & Conceptual Pitfalls
Students often use the net displacement per cycle (2 m in 8 s) as a constant speed and compute $13/2$ cycles, ignoring that the drunkard can reach the pit during the forward part of a cycle. Always check the piecewise $x$‑$t$ graph and stop the count as soon as $x=13$ m is attained, not only at the end of a complete forward‑backward cycle.
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.
Step-by-Step Verified Solution
**Answer:** The drunkard takes 37 s to fall into the pit.
**Step-by-Step Explanation:**
Each cycle: 5 steps forward, 3 steps backward. Net displacement per cycle = 5 - 3 = 2 m. Time per cycle = 8 s.
After 4 cycles (32 s): net displacement = 8 m. [citation:1][citation:11]
In the next forward phase of 5 steps: step 1 (33 s) _ 9 m, step 2 (34 s) _ 10 m, step 3 (35 s) _ 11 m, step 4 (36 s) _ 12 m, step 5 (37 s) _ 13 m and falls into the pit.
Total time = 37 s. [citation:11]
All Questions in this Chapter
Related Learning Resources & Exam Preparation
Chapter Practice Bank
Practice all topic-wise questions from MOTION IN A STRAIGHT LINE
Reference Textbook Solutions
Explore complete chapter exercises for NCERT Physics - Part 1
Competitive PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring