Question : A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward and so on. Each step is 1 m long and requires 1 s. How long the drunkard takes to fall in a pit 13 m away from the street.
Doubt by Yashika
OR
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.
Doubt by Krish
Solution :
Distance covered by the drunkard so that he should fall into the pit = 13 m
Total distance covered by drunkard in first 5 sec = 5 m
Total distance covered by drunkard in first 8 sec (5+3) = 5-3 = 2 m
Total distance covered by drunkard in first 13 sec (8+5) = 2+5 = 7 m
Total distance covered by drunkard in first 16 sec (13+3) = 7-3 = 4 m
Total distance covered by drunkard in first 21 sec (16+5) = 4+5 = 9 m
Total distance covered by drunkard in first 24 sec (21+3) = 9-3 = 6 m
Total distance covered by drunkard in first 29 sec (24+5) = 6+5 = 11 m
Total distance covered by drunkard in first 32 sec (29+3) = 11-3 = 8 m
Total distance covered by drunkard in first 37 sec (32+5) = 8+5 = 13 m
Therefore, the distance of 13 m is coved in first 37 seconds. Hence, the drunkard will fall into the pit after 37 seconds after the starting.
Graphical Solution :
From the graph also it is clear that 13m distance is being covered in 37 seconds. So, the drunkard will fall into the pit after 37 seconds.
Video Solution :