A particle is moving with velocity $\overrightarrow{\mathrm{v}}=\mathrm{K}(\mathrm{y}…

A particle is moving with velocity $\overrightarrow{\mathrm{v}}=\mathrm{K}(\mathrm{y} \hat{\mathrm{i}}+\mathrm{x} \hat{\mathrm{j}})$, where $\mathrm{K}$ is a constant. The general equation for its path is
  1. $y=x^2+$ constant
  2. $y^2=x+$ constant
  3. $x y=$ constant
  4. $y^2=x^2+$ constant

Solution

$\vec{v}=K y \hat{i}+K x \hat{j}$ $\frac{d x}{d t}=K y, \quad \frac{d y}{d t}=K x$ $\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{K x}{K y}$ $y d y=x d x$ $y^2=x^2+c$

Asked in: JEE Main 2010

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