The coordinates of a particle moving in a pla | vedclass

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(D) $x = a \cos(pt)$ and $y = b \sin(pt)$ (given).$	herefore \cos(pt) = x/a$ and $\sin(pt) = y/b$.By squaring and adding:$\cos^2(pt) + \sin^2(pt) = \

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Both $(a)$ and $(b)$.

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(D) $x = a \cos(pt)$ and $y = b \sin(pt)$ (given).$	herefore \cos(pt) = x/a$ and $\sin(pt) = y/b$.By squaring and adding:$\cos^2(pt) + \sin^2(pt) = \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Hence,the path of the particle is an ellipse.Now,differentiating $x$ and $y$ with respect to time:$v_x = \frac{dx}{dt} = -ap \sin(pt)$ and $v_y = \frac{dy}{dt} = bp \cos(pt)$.$\vec{v} = -ap \sin(pt) \hat{i} + bp \cos(pt) \hat{j}$.Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -ap^2 \cos(pt) \hat{i} - bp^2 \sin(pt) \hat{j}$.At $t = \frac{\pi}{2p}$:$\vec{v} = -ap \sin(\pi/2) \hat{i} + bp \cos(\pi/2) \hat{j} = -ap \hat{i}$.$\vec{a} = -ap^2 \cos(\pi/2) \hat{i} - bp^2 \sin(\pi/2) \hat{j} = -bp^2 \hat{j}$.Since $\vec{v} \cdot \vec{a} = (-ap \hat{i}) \cdot (-bp^2 \hat{j}) = 0$,the velocity and acceleration are perpendicular (normal) to each other at $t = \frac{\pi}{2p}$.

The coordinates of a particle moving in a plane are given by $x = a \cos(pt)$ and $y = b \sin(pt)$,where $a, b (b < a)$ and $p$ are positive constants of appropriate dimensions. Then:

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