A particle is moving with a velocity v→$$=kyi^+xj^$$ where k is constant. The general equation for its path is

Correct option is 2y2=x2+ constant

Questions

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

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y=x2+ constant

y2=x2+ constant

y2=x+ constant

xy=constant

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detailed solution

Correct option is B

v→=kyi^+xj^ ⇒vxi+vyj=kyi^+xj^ ⇒dxdti+dydtj=kyi^+xj^ comparing, ⇒dxdt=ky; dydt=kx divide one by the other above equations dxdtdydt=kykx ⇒dxdy=yx ⇒xdx=ydy integrating ∫xdx=∫ydy ⇒y2=x2+c

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A particle is moving with a velocity v→=kyi^+xj^ where k is constant. The general equation for its path is

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free study material

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