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(B) Given,$v = a\sqrt{s}$.Tangential acceleration $a_t = v \frac{dv}{ds}$.Since $\frac{dv}{ds} = a \cdot \frac{1}{2\sqrt{s}} = \frac{a}{2\sqrt{s}}$,we

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$	an \alpha = \frac{2s}{R}$

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(B) Given,$v = a\sqrt{s}$.Tangential acceleration $a_t = v \frac{dv}{ds}$.Since $\frac{dv}{ds} = a \cdot \frac{1}{2\sqrt{s}} = \frac{a}{2\sqrt{s}}$,we have $a_t = (a\sqrt{s}) \cdot (\frac{a}{2\sqrt{s}}) = \frac{a^2}{2}$.Centripetal acceleration $a_c = \frac{v^2}{R} = \frac{(a\sqrt{s})^2}{R} = \frac{a^2 s}{R}$.The angle $\alpha$ between the total acceleration vector and the velocity vector is given by $	an \alpha = \frac{a_c}{a_t}$.Substituting the values,$	an \alpha = \frac{a^2 s / R}{a^2 / 2} = \frac{2s}{R}$.

$A$ particle moves along an arc of a circle of radius $R$. Its velocity depends on the distance covered as $v = a\sqrt{s}$,where $a$ is a constant. Then the angle $\alpha$ between the vector of the total acceleration and the vector of velocity as a function of $s$ will be:

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