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(D) For a particle moving along a curve,the acceleration vector $\vec{a}$ is given by $\vec{a} = \frac{dv}{dt} \hat{t} + \frac{v^2}{R} \hat{n}$,where

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Both $(B)$ and $(C)$.

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(D) For a particle moving along a curve,the acceleration vector $\vec{a}$ is given by $\vec{a} = \frac{dv}{dt} \hat{t} + \frac{v^2}{R} \hat{n}$,where $v$ is the speed,$R$ is the radius of curvature,$\hat{t}$ is the unit tangent vector,and $\hat{n}$ is the unit normal vector.If the speed is constant,$\frac{dv}{dt} = 0$,so $\vec{a} = \frac{v^2}{R} \hat{n}$. This acceleration is directed towards the center of curvature (normal direction),not along the tangent. Thus,option $(B)$ is correct.Since $\vec{a} = \frac{v^2}{R} \hat{n}$,the magnitude of acceleration is $|\vec{a}| = \frac{v^2}{R} = v^2 \kappa$,where $\kappa = \frac{1}{R}$ is the curvature. Thus,the magnitude of acceleration is proportional to the curvature $\kappa$. Thus,option $(C)$ is correct.Therefore,both $(B)$ and $(C)$ are correct.

$A$ particle is moving along a curve. Then

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