Question

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

• A. 12 m
• B. 15 m
• C. 17.32 m
• D. 14.14 m

Answer 1

Answer: (C)

17.32 m

Answer 2

The man's movement forms a regular hexagon, meaning after 6 segments (60° turn each, total 360°), the displacement is zero.

"Just at the start of the 8th turn" usually implies the displacement after 7 completed segments. In this case, the displacement would be that of the 7th segment, which is 10 m (as the 7th segment is in the same direction as the 1st segment, starting from the origin after the 6th segment completed the hexagon). However, 10 m is not an option.

If "just at the start of the 8th turn" is interpreted as the displacement after 8 completed segments, then the total displacement is the vector sum of the 7th and 8th segments (since the first 6 sum to zero).

The 7th segment is $10 \hat{i}$.

The 8th segment is $10 (\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j}) = 5 \hat{i} + 5\sqrt{3} \hat{j}$.

The total displacement is $(10 \hat{i}) + (5 \hat{i} + 5\sqrt{3} \hat{j}) = 15 \hat{i} + 5\sqrt{3} \hat{j}$.

The magnitude is $\sqrt{15^2 + (5\sqrt{3})^2} = \sqrt{225 + 75} = \sqrt{300} = 10\sqrt{3} \approx 17.32$ m.