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${}^{\circ }$?

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

Answer 1

Answer: (C)

r

Let the partical is moving from point $A$ to $B$
Displacement of the partical $=AB$

According to cosine formula
$cos{60}^{\circ }$ = $\frac{r^2+r^2-x^2}{2r^2}$
$2r^{2}cos{60}^{\circ }$= $2r^2-x^2$
$x^2$ = $2r^{2}cos{60}^{\circ }$ = $2r^2\left[2si{n}^2{30}^{\circ }\right]$
$\therefore x=r$
Displacement $AB=x=r$