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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
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The correct answer is:
$y^2=x^2+$ constant
$y^2=x^2+$ constant
$\vec{v}=K y \hat{i}+K x \hat{j}$
$\frac{d x}{d t}=K y, \quad \frac{d y}{d t}=K x$
$\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{K x}{K y}$
$y d y=x d x$
$y^2=x^2+c$
$\frac{d x}{d t}=K y, \quad \frac{d y}{d t}=K x$
$\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{K x}{K y}$
$y d y=x d x$
$y^2=x^2+c$
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