A man moves in an open field such that after moving 10 ,m on a straight line, he makes a sharp turn

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3-2.Motion in Plane

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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 $8^{\text {th }}$ turn is equal to ........$m$

A

$12$

B

$15$

C

$17.32$

D

$14.14$

Solution

(c)

When a man moves $10 m$ before turning by $60^{\circ}$, he makes a hexagon in six steps.

$\therefore$ In eight steps he will trace two additional stegs of hexagon from starting point

$\therefore$ Displacement is $A B$

$=\sqrt{10^2+10^2+2 \times 10 \times 10 \cos 60}$

$=\sqrt{100+100+200\left(\frac{1}{2}\right)}$

$=10 \sqrt{3}=17.32\,m$

Standard 11

Physics

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