A particle is in uniform circular motion. The | vedclass

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(C) The equation of the trajectory is $(x-2)^2 + y^2 = 25$. This represents a circle with center at $(2, 0)$ and radius $R = 5 	ext{ m}$.In uniform c

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$0.8 \hat{j}$

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(C) The equation of the trajectory is $(x-2)^2 + y^2 = 25$. This represents a circle with center at $(2, 0)$ and radius $R = 5 	ext{ m}$.In uniform circular motion,the acceleration is purely centripetal,directed towards the center of the circle.The magnitude of centripetal acceleration is $a_c = \frac{v^2}{R}$.Given $v = 2 	ext{ m/s}$ and $R = 5 	ext{ m}$,we have $a_c = \frac{2^2}{5} = \frac{4}{5} = 0.8 	ext{ m/s}^2$.The particle attains the lowest $y$ coordinate at the point $(2, -5)$.At this point,the center of the circle $(2, 0)$ is directly above the particle (along the positive $y$-axis).Therefore,the centripetal acceleration vector is directed along the positive $y$-axis,which is $0.8 \hat{j} 	ext{ m/s}^2$.

$A$ particle is in uniform circular motion. The equation of its trajectory is given by $(x-2)^2+y^2=25$,where $x$ and $y$ are in meters. The speed of the particle is $2 \text{ m/s}$. When the particle attains the lowest $y$ coordinate,the acceleration of the particle is (in $\text{m/s}^2$):

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