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 graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
The drunkard takes 37 seconds to fall in the pit 13 m away from the start.
Explanation
This problem involves analyzing periodic motion with forward and backward segments. The key insight is that each complete cycle (5 forward + 3 backward steps) results in a net displacement of 2 m in 8 seconds. The student must recognize that the drunkard reaches 13 m during the forward motion of the 5th cycle, not after completing it. The graphical method involves plotting the sawtooth pattern and reading the time coordinate where x = 13 m. The analytical method uses the formula: after n complete cycles, position = 2n m, and additional forward steps add to this position.
Solution Steps
Step 1: Analyze one complete cycle - Forward: 5 steps = 5 m in 5 s; Backward: 3 steps = 3 m in 3 s. Net displacement per cycle = 5 - 3 = 2 m. Time per cycle = 5 + 3 = 8 s.
Step 2: Calculate position after each cycle - After 1st cycle (8 s): 2 m; After 2nd cycle (16 s): 4 m; After 3rd cycle (24 s): 6 m; After 4th cycle (32 s): 8 m.
Step 3: Apply analytical method - After n complete cycles, position = 2n m. We need position = 13 m. Using formula: 2n + k = 13, where k = additional forward steps (k 5). For n = 4: k = 13 - 8 = 5 (valid).
Step 4: Calculate total time - Time = 8n + k = 8(4) + 5 = 32 + 5 = 37 s.
Step 5: Graphical verification - Plot x-t graph with points: (0,0) → (5,5) → (8,2) → (13,7) → (16,4) → (21,9) → (24,6) → (29,11) → (32,8) → (37,13). At point (37 s, 13 m), the drunkard falls in the pit.