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Standard VII

Physics

Displacement

A drunkard ta...

Question

A drunkard takes a step of $1 m$ in $1 s$ . He takes 5 steps forward and 3 steps backward and so on. The time taken by him to fall in a pit $13 m$ away from the start is :

26 s

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31 s

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37 s

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41 s

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Solution

The correct option is C $37 s$
The man takes 5 steps forward and 3 steps backwards, distance from pit is 13 meters and distance of 1 step is 1 meter.
$5 m ⟶ 2 m 7 m ⟶ 4 m 9 m ⟶ 6 m 11 m ⟶ 8 m 13 m$

Each step needs $8 s e c$ , 5 in forward direction and 3 in backward direction.
Now, he reaches the distance of $8 m$ in $32 s e c$ .
Thus, he reaches the distance of $13 m$ in $32 + 5 = 37 s e c$ .

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