Question

A man moves in an open field such that after moving 10 m in a straight line, he makes a sharp turn of 60° to his left. The total displacement of the man just after 7 such turns is

Answer 1

10 m

Answer 2

Solution:

Notice that when a man walks 10 m and turns 60° to his left repeatedly, his path follows the edges of a regular hexagon. After 6 such segments the hexagon closes and he returns to his starting point. Thus, upon completing 6 turns he is back at the origin. One more segment (after the 7th turn) will be along the direction of the first side. Hence the net displacement is just that extra 10 m.

Explanation (minimal):

1. The man’s path forms a regular hexagon because the external turn is 60°.
2. After 6 segments (i.e. 6 turns) the displacement is zero.
3. The 7th segment adds 10 m in the direction of the first side.
4. Therefore, the net displacement = 10 m.