A particle moves along the curve \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\), with constant speed \(v\). Express…
- \(\frac{(\mp 9 y \hat{i} \pm 4 x \hat{j}) v}{\sqrt{16 x^{2}+81 y^{2}}}\)
- \(\frac{(\mp 9 y \hat{i} \pm 7 x \hat{j}) v}{\sqrt{16 x^{2}+81 y^{2}}}\)
- \(\frac{(\mp 9 y \hat{i} \pm 9 x \hat{j}) v}{\sqrt{16 x^{2}+81 y^{2}}}\)
- \(\frac{(\mp 9 y \hat{i} \pm 5 x \hat{j}) v}{\sqrt{16 x^{2}+81 y^{2}}}\)
Solution
Differentiating both sides w.r.t. time \(\Rightarrow \quad \frac{1}{9} \frac{d x^{2}}{d t}+\frac{1}{4} \frac{d y^{2}}{d t}=\frac{d(1)}{d t}\)
\(\Rightarrow \quad \frac{1}{9} 2 x \cdot \frac{d x}{d t}+\frac{1}{4} \cdot 2 y \cdot \frac{d y}{d t}=0\)
\(\Rightarrow \quad \frac{2 x}{9}\left[\frac{d x}{d t}\right]+\frac{2 y}{4}\left[\frac{d y}{d t}\right]=0\)
\(\Rightarrow \quad v_{x}=-\frac{9 y}{4 x} v_{y} \quad\left[\frac{d x}{d t}=v_{x} ; \frac{d y}{d t}=v_{y}\right] \ldots(\mathrm{i})\)
As particle moves with constant velocity, then
\(v_{x}^{2}+v_{y}^{2}=v^{2} \Rightarrow\left(-\frac{9}{4} \frac{y}{x} v_{y}\right)^{2}+v_{y}^{2}=v^{2}\)
\(\Rightarrow \quad v_{y}^{2}=\frac{16 x^{2} v^{2}}{16 x^{2}+81 y^{2}} \Rightarrow v_{y}=\frac{\pm 4 x v}{\sqrt{16 x^{2}+81 y^{2}}}\) ...(ii)
From (i) and (ii), we get \(v_{x}=\frac{\mp 9 y v}{\sqrt{16 x^{2}+81 y^{2}}}\)
So, velocity is given by \(\vec{v}=v_{x} \hat{i}+v_{y} \hat{j}=\frac{(\mp 9 y \hat{i} \pm 4 x \hat{j}) v}{\sqrt{16 x^{2}+81 y^{2}}}\) ^
Asked in: JEE Mains - Motion In One Dimension - Test 2