Question

A particle is moving with velocity $\overrightarrow{v}=K(y\hat{i}+x\hat{j})$, where $K$ is a constant. The general equation for its path is :

• A. $y=x^2+$ constant
• B. $y^2=x+$ constant
• C. $xy=$ constant
• D. $y^2=x^2+$ constant

Answer 1

The correct option is D $y^2=x^2+$ constant

The expression for the velocity of particle is:

$\overrightarrow{v}=$ $Ky$ $\hat{i}+Kx\hat{j}$
From the equation, we get

$\frac{dx}{dt}=$ $Ky$, and $\frac{dy}{dt}=$ $Kx$
From the above equation, $\frac{dy}{dx}$ is obtained as:

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{Kx}{Ky}$

$ydy=xdx$

Now, integrating on both side:

$y\int dy=x\int dx$

$⟹y^2=x^2+c$