Motion in a Straight Line
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Question : 4
Total: 27
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.
Solution:
The x-t graph of drunkard is shown in figure.
∴ Time taken to move 5 steps = 5 s.
5 steps (i.e. 5 m) forward and 3 steps (i.e. 3 m) backward means that the net distance covered by him in first 8 steps i.e. in 8 s = 5 m – 3 m = 2 m.
Distance covered by him in first 16 steps or 16 s = 2 + 2 = 4 m.
Distance covered by the drunkard in first 24 i.e. 24 steps = 2 + 2 + 2 = 6 m and distance covered in 32 steps i.e. 32 sec = 8 m.
∴ Distance covered in first 37 steps = 8 + 5 = 13 m.
Distance of the pit from the start = 13 m.
∴ Total time taken by the drunkard to fall in the pit = 37 s.
Since 1 step requires 1 s of time, so we arrive at the same result from the graph shown.
Length of each step = 1 m, time taken for each step = 1 s.
∴ Time taken to move 5 steps = 5 s.
5 steps (i.e. 5 m) forward and 3 steps (i.e. 3 m) backward means that the net distance covered by him in first 8 steps i.e. in 8 s = 5 m – 3 m = 2 m.
Distance covered by him in first 16 steps or 16 s = 2 + 2 = 4 m.
Distance covered by the drunkard in first 24 i.e. 24 steps = 2 + 2 + 2 = 6 m and distance covered in 32 steps i.e. 32 sec = 8 m.
∴ Distance covered in first 37 steps = 8 + 5 = 13 m.
Distance of the pit from the start = 13 m.
∴ Total time taken by the drunkard to fall in the pit = 37 s.
Since 1 step requires 1 s of time, so we arrive at the same result from the graph shown.
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