JEE MAIN - Physics (2010 - No. 27)
A particle is moving with velocity $$\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$$, where K is a constant. The general equation for its path is
y = x2 + constant
y2 = x + constant
xy = constant
y2 = x2 + constant
Explanation
$$\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$$ ........(1)
Also $$\overrightarrow v = {v_x}\widehat i + {v_y}\widehat j$$
$$\overrightarrow v = {{dx} \over {dt}}\widehat i + {{dy} \over {dt}}\widehat j$$ ........(2)
Equating (1) and (2), we get
$${{dx} \over {dt}} = ky\,\,\,\,\,\,$$ .......(3)
and $$\,\,\,\,\,{{dy} \over {dt}} = kx$$ ......(4)
Dividing (3) and (4), we get
$${{dy} \over {dx}} = {x \over y} $$
$$\Rightarrow ydy = xdx$$
Integrating both sides of above equation, we get
$$\int {ydy} = \int {xdx} $$
$$ \Rightarrow {y^2} = {x^2} + $$ constant
Also $$\overrightarrow v = {v_x}\widehat i + {v_y}\widehat j$$
$$\overrightarrow v = {{dx} \over {dt}}\widehat i + {{dy} \over {dt}}\widehat j$$ ........(2)
Equating (1) and (2), we get
$${{dx} \over {dt}} = ky\,\,\,\,\,\,$$ .......(3)
and $$\,\,\,\,\,{{dy} \over {dt}} = kx$$ ......(4)
Dividing (3) and (4), we get
$${{dy} \over {dx}} = {x \over y} $$
$$\Rightarrow ydy = xdx$$
Integrating both sides of above equation, we get
$$\int {ydy} = \int {xdx} $$
$$ \Rightarrow {y^2} = {x^2} + $$ constant
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