AP EAMCET · PHYSICS · Motion In Two Dimensions
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
- A \(\tan \theta=\sqrt{\frac{S}{R}}\)
- B \(\tan \theta=\sqrt{\frac{S}{2R}}\)
- C \(\tan \theta={\frac{S}{2R}}\)
- D \(\tan \theta={\frac{2S}{R}}\)
Answer & Solution
Correct Answer
(D) \(\tan \theta={\frac{2S}{R}}\)
Step-by-step Solution
Detailed explanation
\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…
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