The position vector of a particle vec{R} as a | vedclass

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(D) Given position vector: $\vec{R} = 4\sin(2\pi t)\hat{i} + 4\cos(2\pi t)\hat{j}$.Since $x = 4\sin(2\pi t)$ and $y = 4\cos(2\pi t)$,then $x^2 + y^2 =

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Magnitude of the velocity of the particle is $8 \ m/s$.

Vedclass · Answer

(D) Given position vector: $\vec{R} = 4\sin(2\pi t)\hat{i} + 4\cos(2\pi t)\hat{j}$.Since $x = 4\sin(2\pi t)$ and $y = 4\cos(2\pi t)$,then $x^2 + y^2 = 16(\sin^2(2\pi t) + \cos^2(2\pi t)) = 16$. This represents a circle of radius $4 \ m$.Velocity $\vec{v} = \frac{d\vec{R}}{dt} = 8\pi\cos(2\pi t)\hat{i} - 8\pi\sin(2\pi t)\hat{j}$.Magnitude of velocity $|\vec{v}| = \sqrt{(8\pi\cos(2\pi t))^2 + (-8\pi\sin(2\pi t))^2} = 8\pi \ m/s$.Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -16\pi^2\sin(2\pi t)\hat{i} - 16\pi^2\cos(2\pi t)\hat{j} = -4\pi^2\vec{R}$.Since $\vec{a} = -4\pi^2\vec{R}$,the acceleration is directed towards the origin (along $-\vec{R}$).For uniform circular motion,$a = \frac{V^2}{R}$. Here $V = 8\pi$ and $R = 4$,so $a = \frac{(8\pi)^2}{4} = 16\pi^2$. This matches the magnitude of $\vec{a}$.Statement $(D)$ claims the magnitude of velocity is $8 \ m/s$,but it is $8\pi \ m/s$. Thus,$(D)$ is the wrong statement.

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