If a particle is moving with velocity v - y -2 x - then the locus of its path is A- y -2 x - cB- 2 y - x - cC- y2-2 x2-cD- 2 y 2- x 2- c

The correct option is C y^2=2x^2+cV_x=dx/dt=y, V_y=dy/dt=2x dy/dx=dy/dt/dx/dt=2x/y ydy=2xdx integrating both sides y^2/2=2x^2/2+c y^2=2x^2+c ...

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Standard XII

Physics

Differentiation

If a particle...

Question

If a particle is moving with velocity $v = y^i + 2 x^j$ , then the locus of its path is

y = 2 x + c

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2 y = x + c

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y^{2} = 2 x^{2} + c

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2 y^{2} = x^{2} + c

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Solution

The correct option is C $y^{2} = 2 x^{2} + c$
$V_{x} = \frac{d x}{d t} = y, V_{y} = \frac{d y}{d t} = 2 x$
$\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{2 x}{y}$
$y d y = 2 x d x$
integrating both sides
$\frac{y^{2}}{2} = \frac{2 x^{2}}{2} + c$
$y^{2} = 2 x^{2} + c$

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If a particle is moving with velocity v=y^i+2x^j, then the locus of its path is

The correct option is C y2=2x2+cVx=dxdt=y,Vy=dydt=2x dydx=dy/dtdx/dt=2xy ydy=2xdx integrating both sides y22=2x22+c y2=2x2+c

If a particle is moving with velocity $v = y^i + 2 x^j$ , then the locus of its path is

The correct option is C $y^{2} = 2 x^{2} + c$
$V_{x} = \frac{d x}{d t} = y, V_{y} = \frac{d y}{d t} = 2 x$
$\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{2 x}{y}$
$y d y = 2 x d x$
integrating both sides
$\frac{y^{2}}{2} = \frac{2 x^{2}}{2} + c$
$y^{2} = 2 x^{2} + c$