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 $ heta$ is the angle between the total acceleration and tangential acceleration, then

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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 $	heta$ is the angle between the total acceleration and tangential acceleration, then

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } \because \quad 	an 	heta=\frac{a_{	ext {radial }}}{a_{	ext {tangential }}} \ & =\frac{\frac{V^2}{R}}{a_{	ext {tangential }}} \ & 	ext { Givn } V=K \sqrt{s} \ & \Rightarrow 	an 	heta=\frac{1}{a_{	ext {tangential }}}\left[\frac{1}{R} 	imes K^2 sight] \ & \Rightarrow 	an 	heta=\frac{1}{a_{	ext {tangential }}}\left[\frac{K^2 s}{R}ight]\end{aligned}$ $\begin{aligned} & 	ext { Now, } a_{	ext {tangential }}=\frac{d v}{d t}=\frac{d}{d t}=[K \sqrt{s}] \ & 	ext { or } a_{	ext {tangential }}=K 	imes \frac{1}{2 \sqrt{s}} 	imes \frac{d s}{d t} \ & 	ext { or } a_{	ext {tangential }}=\frac{K}{2 \sqrt{s}} 	imes V \ & =\frac{K}{2 \sqrt{s}} 	imes K \sqrt{s}=\frac{K^2}{2} \ & 	herefore 	an 	heta=\frac{\frac{K^2 s}{R}}{\frac{K^2}{2}} \Rightarrow 	an 	heta=\frac{2 s}{R} \end{aligned}$

A point object moves along an arc of a circle of radius $R$. Its velocity depends upon the distance covered…

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