A particle is moving with velocity vec v = K( | vedclass

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(A) Given the velocity vector $\vec v = K(y\hat i + x\hat j)$.We know that $\vec v = v_x \hat i + v_y \hat j = \frac{dx}{dt} \hat i + \frac{dy}{dt} \h

Vedclass · Accepted Answer

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

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(A) Given the velocity vector $\vec v = K(y\hat i + x\hat j)$.We know that $\vec v = v_x \hat i + v_y \hat j = \frac{dx}{dt} \hat i + \frac{dy}{dt} \hat j$.Comparing the components,we get $\frac{dx}{dt} = Ky$ and $\frac{dy}{dt} = Kx$.To find the path of the particle,we divide the two equations: $\frac{dy/dt}{dx/dt} = \frac{Kx}{Ky}$.This simplifies to $\frac{dy}{dx} = \frac{x}{y}$.Rearranging the terms,we get $y \, dy = x \, dx$.Integrating both sides,we obtain $\int y \, dy = \int x \, dx$.This results in $\frac{y^2}{2} = \frac{x^2}{2} + C'$,where $C'$ is a constant.Multiplying by $2$,we get $y^2 = x^2 + C$,where $C = 2C'$ is another constant.Thus,the equation of the path is $y^2 = x^2 + 	ext{constant}$.

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

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