Q. A particle is moving with velocity $\mathbf{v} = k\left( y\widehat{\dot{\mathbf{i}}} + x\ \widehat{\dot{\mathbf{j}}} \right),$ where $k\ $is a constant. The general equation for its path is
A) $y = x^{2} + constant$
B) $y^{2} = x + constant$
C) $xy = constant$
D) $y^{2} = x^{2}constant$
Explanation
(d) $v = ky\widehat{\dot{\mathbf{i}}} + kx\ \widehat{\dot{\mathbf{j}}}$
$\frac{dx}{dt} = ky,\ \frac{dy}{dt} = kx$
$\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{kx}{ky}$
$ydy = xdx$
$y^{2} = x^{2} +$c
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.