Q. A particle is moving with velocity $\overrightarrow{v} = K(y\widehat{i} + x\widehat{j})$, where $K$ is a constant. The general equation for its path is
A) $y^{2} = x^{2} + \ $constant
B) $y = x^{2} + \ $constant
C) $y^{2} = x + \ $constant
D) $xy =$ constant
Explanation
(a) $v = K(y\widehat{i} + x\widehat{j})$
$v_{x} = Ky$
$\frac{dx}{dt} = Ky$
Similarly, $\frac{dy}{dt} = Kx$
Hence $\frac{dy}{dx} = \frac{x}{y}$
$\Rightarrow \ y\ dy = x\ dx$ , by integrating
$y^{2} = x^{2} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.