A particle is moving with a velocity v = K(yi + xj), where K is a constant. The general equation for its path is: (a) y = x2 + constant (b) y2 = x + constant (c) y2 = x2 + constant (d) xy = constant

Answer: (c). dx/dt = Ky, dy/dt = Kx ⇒ dy/dx = x/y y dy = x dx ⇒ y² = x² + constant So the answer is (c).

A particle is moving with a velocity v = K(yi + xj), where K is a… | JEE Main 2019 (9 Jan, Shift 1)

Q13JEE Main 2019 · 9 JAN · SHIFT 1Medium

Motion in a Plane (2-D Kinematics)

A particle is moving with a velocity

$\vec{v} = K(y\hat{i} + x\hat{j}),$ where K is a constant.

The general equation for its path is:

Show answer and solution

Answer: (c)

$\frac{dx}{dt} = Ky$ , $\frac{dy}{dt} = Kx \Rightarrow \frac{dy}{dx} = \frac{x}{y}$
$y\,dy = x\,dx \Rightarrow y^2 = x^2 + \text{constant}$
So the answer is (c).