A particle is moving on a circular path of radius r with uniform speed v . displacement of particle

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3-2.Motion in Plane

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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$2r$

Solution

According to cosine formula
$\cos 60^{\circ}=\frac{r^{2}+r^{2}-x^{2}}{2 r^{2}}$
$2 \mathrm{r}^{2} \cos 60^{\circ}=2 \mathrm{r}^{2}-\mathrm{x}^{2}$
$\mathrm{x}^{2}=2 \mathrm{r}^{2}-2 \mathrm{r}^{2} \cos 60^{\circ}=2 \mathrm{r}^{2}\left[1-\cos 60^{\circ}\right]$
$=2 \mathrm{r}^{2}\left[2 \sin ^{2} 30^{\circ}\right]=\mathrm{r}^{2}$
$\therefore x=r$
Displacement $\mathrm{AB}=\mathrm{x}=\mathrm{r}$

Standard 11

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(JEE MAIN-2024)

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 $60^o$ ?

Solution

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