A drunkard walking in a narrow lane takes 5 s | vedclass

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Question

(B) Distance covered in $1$ step $= 1\; m$.Time taken for $1$ step $= 1\; s$.Time taken to move $5\; m$ forward $= 5\; s$.Time taken to move $3\; m$ b

Vedclass · Accepted Answer

$37$

Vedclass · Answer

(B) Distance covered in $1$ step $= 1\; m$.Time taken for $1$ step $= 1\; s$.Time taken to move $5\; m$ forward $= 5\; s$.Time taken to move $3\; m$ backward $= 3\; s$.Net distance covered in one cycle $= 5 - 3 = 2\; m$.Net time taken for one cycle $= 5 + 3 = 8\; s$.To reach the pit at $13\; m$,the drunkard must reach $8\; m$ first,because in the next forward movement of $5\; m$,he will reach $8 + 5 = 13\; m$ and fall into the pit.Time taken to cover $8\; m$ (which is $4$ cycles) $= 4 	imes 8\; s = 32\; s$.After $32\; s$,the drunkard is at $8\; m$. In the next $5\; s$,he takes $5$ steps forward to reach $8 + 5 = 13\; m$.Total time taken $= 32\; s + 5\; s = 37\; s$.

Column $I$	Column $II$
$(A)$ Change in velocity	$(p)$ $-5/3\,SI \text{ unit}$
$(B)$ Average acceleration	$(q)$ $-20\,SI \text{ unit}$
$(C)$ Total displacement	$(r)$ $-10\,SI \text{ unit}$
$(D)$ Acceleration at $t=3\,s$	$(s)$ $-5\,SI \text{ unit}$

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