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 after 8 such turn is equal to

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 after 8 such turn is equal to (A) 12 m (B) 15 m (C) 17.32 m (D) 14.14 m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7ddca780801ccd6bc44294ca2cf4b9f8-.png" /> After 8 such turns object is at ' B '. Displacement =A B Two vectors are at 60° √102+102+2 × 102 × (1/2) =10 √3 m ⇒ 17.32 m = AB

Q. A man moves in an open field such that after moving $10 m$ on a straight line, he makes a sharp turn of $6 0^{\circ}$ to his left. The total displacement after $8$ such turn is equal to

12 m

15 m

17.32 m

14.14 m

After 8 such turns object is at ' $B$ '.
Displacement $= A B$
Two vectors are at $6 0^{\circ}$
$1 0^{2} + 1 0^{2} + 2 \times 1 0^{2} \times \frac{1}{2}$
$= 103 m$
$\Rightarrow 17.32 m = A B$

Q. A man moves in an open field such that after moving 10m on a straight line, he makes a sharp turn of 60∘ to his left. The total displacement after 8 such turn is equal to