Kinematics

JEE Main level, single correct, medium

274 of 308

Question 274 of 308

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

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Answer: (D) y2=x2+y^2=x^2+ constant

v⃗=Kyi^+Kxj^\vec{v}=K y \hat{i}+K x \hat{j}
dxdt=Ky,dydt=Kx\frac{d x}{d t}=K y, \quad \frac{d y}{d t}=K x
dydx=dydt×dtdx=KxKy\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=\frac{K x}{K y}
ydy=xdxy d y=x d x
y2=x2+cy^2=x^2+c

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