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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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The correct answer is:
$\tan \theta={\frac{2S}{R}}$
$\begin{aligned} & \text { } \because \quad \tan \theta=\frac{a_{\text {radial }}}{a_{\text {tangential }}} \\ & =\frac{\frac{V^2}{R}}{a_{\text {tangential }}} \\ & \text { Givn } V=K \sqrt{s} \\ & \Rightarrow \tan \theta=\frac{1}{a_{\text {tangential }}}\left[\frac{1}{R} \times K^2 s\right] \\ & \Rightarrow \tan \theta=\frac{1}{a_{\text {tangential }}}\left[\frac{K^2 s}{R}\right]\end{aligned}$
$\begin{aligned}
& \text { Now, } a_{\text {tangential }}=\frac{d v}{d t}=\frac{d}{d t}=[K \sqrt{s}] \\
& \text { or } a_{\text {tangential }}=K \times \frac{1}{2 \sqrt{s}} \times \frac{d s}{d t} \\
& \text { or } a_{\text {tangential }}=\frac{K}{2 \sqrt{s}} \times V \\
& =\frac{K}{2 \sqrt{s}} \times K \sqrt{s}=\frac{K^2}{2} \\
& \therefore \tan \theta=\frac{\frac{K^2 s}{R}}{\frac{K^2}{2}} \Rightarrow \tan \theta=\frac{2 s}{R}
\end{aligned}$
$\begin{aligned}
& \text { Now, } a_{\text {tangential }}=\frac{d v}{d t}=\frac{d}{d t}=[K \sqrt{s}] \\
& \text { or } a_{\text {tangential }}=K \times \frac{1}{2 \sqrt{s}} \times \frac{d s}{d t} \\
& \text { or } a_{\text {tangential }}=\frac{K}{2 \sqrt{s}} \times V \\
& =\frac{K}{2 \sqrt{s}} \times K \sqrt{s}=\frac{K^2}{2} \\
& \therefore \tan \theta=\frac{\frac{K^2 s}{R}}{\frac{K^2}{2}} \Rightarrow \tan \theta=\frac{2 s}{R}
\end{aligned}$
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