A particle is moving on a circular path of ra | vedclass

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Question

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

Vedclass · Accepted Answer

$r$

Vedclass · Answer

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

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 $60^o$ ?

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