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(C) The man moves $10 \, m$ and turns $60^{\circ}$ to the left. This path forms the sides of a regular hexagon.After $6$ turns,the man returns to the

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(C) The man moves $10 \, m$ and turns $60^{\circ}$ to the left. This path forms the sides of a regular hexagon.After $6$ turns,the man returns to the starting point because $6 	imes 60^{\circ} = 360^{\circ}$.At the start of the $7^{	ext{th}}$ turn,he has completed $6$ segments and is at the origin. He then moves $10 \, m$ for the $7^{	ext{th}}$ segment.At the start of the $8^{	ext{th}}$ turn,he has completed $7$ segments. The displacement is the distance between the starting point and the end of the $7^{	ext{th}}$ segment.Since the first $6$ segments form a closed hexagon,the displacement after $7$ segments is simply the length of the $7^{	ext{th}}$ segment,which is $10 \, m$. However,if the question implies the displacement after $7$ segments where each segment is at $60^{\circ}$ to the previous one,the displacement after $n$ segments is $d = 2R \sin(n	heta/2)$. For $n=7$ and $	heta=60^{\circ}$,$d = 2(10) \sin(7 	imes 30^{\circ}) = 20 \sin(210^{\circ}) = 20(-0.5) = -10$. The magnitude is $10 \, m$. Wait,re-evaluating: After $7$ segments,the position is $10 \, m$ from the origin at an angle of $60^{\circ}$ from the first segment. The displacement is $10 \, m$.

Column $I$	Column $II$
$(A)$ Distance travelled in $3\,s$	$(p)$ $-20$ units
$(B)$ Displacement in $1\,s$	$(q)$ $15$ units
$(C)$ Initial acceleration	$(r)$ $25$ units
$(D)$ Velocity at $4\,s$	$(s)$ $-10$ units

$A$ man moves in an open field such that after moving $10 \, m$ on a straight line,he makes a sharp turn of $60^{\circ}$ to his left. The total displacement just at the start of the $8^{\text{th}}$ turn is equal to ........ $m$.

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Column $I$ Column $II$

$(A)$ Distance travelled in $3\,s$ $(p)$ $-20$ units

$(B)$ Displacement in $1\,s$ $(q)$ $15$ units

$(C)$ Initial acceleration $(r)$ $25$ units

$(D)$ Velocity at $4\,s$ $(s)$ $-10$ units

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