(D) The radial (centripetal) acceleration is given by $a_r = \frac{v^2}{R}$.Given $v = K\sqrt{s}$,we have $a_r = \frac{(K\sqrt{s})^2}{R} = \frac{K^2 s

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(D) The radial (centripetal) acceleration is given by $a_r = \frac{v^2}{R}$.Given $v = K\sqrt{s}$,we have $a_r = \frac{(K\sqrt{s})^2}{R} = \frac{K^2 s}{R}$.The tangential acceleration is $a_t = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds}$.Since $v = K s^{1/2}$,then $\frac{dv}{ds} = K \cdot \frac{1}{2} s^{-1/2} = \frac{K}{2\sqrt{s}}$.Thus,$a_t = (K\sqrt{s}) \cdot \left( \frac{K}{2\sqrt{s}} \right) = \frac{K^2}{2}$.The angle $\theta$ between the total acceleration and the tangential acceleration is given by $\tan \theta = \frac{a_r}{a_t}$.Substituting the values,$\tan \theta = \frac{K^2 s / R}{K^2 / 2} = \frac{2s}{R}$.

Class 11 Physics 3-2.Motion in Plane Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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$A$ point object moves along an arc of a circle of radius $R$. Its velocity depends upon the distance covered $s$ as $v=K \sqrt{s}$,where $K$ is a constant. If $\theta$ is the angle between the total acceleration and tangential acceleration,then

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A
$\tan \theta=\sqrt{\frac{s}{R}}$
B
$\tan \theta=\sqrt{\frac{s}{2R}}$
C
$\tan \theta=\frac{s}{2R}$
D
$\tan \theta=\frac{2s}{R}$

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3-2.Motion in Plane

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Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

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$A$ point object moves along an arc of a circle of radius $R$. Its velocity depends upon the distance covered $s$ as $v=K \sqrt{s}$,where $K$ is a constant. If $\theta$ is the angle between the total acceleration and tangential acceleration,then

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For circular motion,if $\vec a_t$,$\vec a_c$,$\vec r$ and $\vec v$ are tangential acceleration,centripetal acceleration,radius vector and velocity respectively,then find the wrong relation.

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