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:

(C) $y^2 = x^2 + \text{constant}$

(A) $y = x^2 + \text{constant}$

(B) $y^2 = x + \text{constant}$

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Kinematics

A particle is moving with a velocity

v = K (y \hat{i} + x \hat{j})

, where K is a constant. The general equation for its path is:

(A) $y = x^{2} + constant$
(B) $y^{2} = x + constant$
(C) $y^{2} = x^{2} + constant$ ✓ Correct
(D) $x y = constant$

Answer: (C)

Solution

Read off the two velocity components from

v = K (y \hat{i} + x \hat{j})

\frac{d x}{d t} = K y

and

\frac{d y}{d t} = K x

. Dividing one by the other removes time and leaves the slope of the path,

\frac{d y}{d x} = \frac{x}{y}

. Separating variables gives

y d y = x d x

, and integrating both sides lands on

\frac{y ^{2}}{2} = \frac{x ^{2}}{2} + C

, that is

y^{2} = x^{2} + constant

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option

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