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JEE Main 2010 (AIEEE) · Motion in a Plane (2-D Kinematics)

Motion in a Plane (2-D Kinematics) question from JEE Main 2010 (AIEEE)

Q22JEE Main 2010 · AIEEEMedium

Motion in a Plane (2-D Kinematics)

A particle is moving with velocityv⃗=k(yi^+xj^)\vec{v} = k\left(y\hat{i} + x\hat{j}\right), where K is a constant. The general equation for its path is

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Answer: (d)

  1. dxdt=ky\frac{dx}{dt} = ky, dydt=kx⇒dydx=xy\frac{dy}{dt} = kx \Rightarrow \frac{dy}{dx} = \frac{x}{y}
  2. y dy=x dx⇒y2=x2+constanty\,dy = x\,dx \Rightarrow y^2 = x^2 + \text{constant}
  3. So the answer is (d).
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