Test Index

Motion in a Plane

Section: Physics

© examsnet.com

Question : 89 of 94

Marks: +1, -0

A particle is moving with velocity

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

K

is a constant. The general equation for its path is

$y=x^2+$ constant
$y^2=x+$ constant
$x y=$ constant
$y^2=x^2+$ constant

Solution:

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

.....(1)

Also

\vec{v}=v_x\hat{i}+v_y\hat{j}

\vec{v}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}

.....(2)

Equating (1) and (2), we get

\frac{dx}{dt}=ky

......(3)

and

\frac{dy}{dt}=kx

.....(4)

Dividing (3) and (4), we get

\frac{dy}{dx}=\frac{x}{y}

\Rightarrow y\,dy=x\,dx

Integrating both sides of above equation, we get

\int y\,dy=\int x\,dx

\Rightarrow y^2=x^2+\text{ constant }

© examsnet.com

Go to Question:

Prev Question

Next Question