Question

A particle is moving on a circular path of radius r with uniform speed v. What is the displacement of the particle after it has described an angle of $60^{o}$?

• A. $r\sqrt{2}$
• B. $r\sqrt{3}$
• C. r
• D. 2r

Answer 1

Answer: (C)

r

Answer 2

According to cosine formula

$\cos 60{^\circ} = \frac{r^{2} + r^{2} - x^{2}}{2r^{2}}$

$2r^{2}\cos 60{^\circ} = 2r^{2} - x^{2}$

$x^{2} = 2r^{2} - 2r^{2}\cos 60{^\circ} = 2r^{2}\lbrack 1 - \cos 60{^\circ}\rbrack$

$= 2r^{2}\lbrack 2\sin^{2}30{^\circ}\rbrack = r^{2}$

$\therefore x = r$

Displacement AB = x = r