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(D) The magnitude of the average velocity is given by the formula: $|\vec{v}_{avg}| = \frac{|\vec{r}_B - \vec{r}_A|}{\Delta t}$.The displacement $|\ve

Vedclass · Accepted Answer

$1.5\sqrt{3} \, m/s$

Vedclass · Answer

(D) The magnitude of the average velocity is given by the formula: $|\vec{v}_{avg}| = \frac{|\vec{r}_B - \vec{r}_A|}{\Delta t}$.The displacement $|\vec{r}_B - \vec{r}_A|$ is the chord length $AB$. In a circle of radius $R$,the chord length for an angle $	heta = 120^{\circ}$ is $2R \sin(	heta/2) = 2R \sin(60^{\circ}) = 2R(\sqrt{3}/2) = R\sqrt{3}$.The time taken $\Delta t$ is the distance along the arc divided by speed $v$. The arc length is $s = R	heta = R(2\pi/3)$.Thus,$\Delta t = s/v = (2\pi R / 3) / \pi = 2R/3$.Therefore,$|\vec{v}_{avg}| = \frac{R\sqrt{3}}{2R/3} = \frac{3\sqrt{3}}{2} = 1.5\sqrt{3} \, m/s$.

As shown in the figure,a particle is moving with a constant speed $v = \pi \, m/s$ in a circular path. Considering its motion from point $A$ to point $B$,where the angle subtended at the center is $120^{\circ}$,the magnitude of the average velocity is:

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