JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 7)
A particle is moving with a velocity
$$\overrightarrow v \, = K(y\widehat i + x\widehat j),$$ where K is a constant.
The general equation for its path is :
$$\overrightarrow v \, = K(y\widehat i + x\widehat j),$$ where K is a constant.
The general equation for its path is :
y = x2 + constant
y2 = x + constant
y2 = x2 + constant
xy = constant
Explanation
Given,
$$\overrightarrow v = K\left( {y\widehat i + x\widehat j} \right)$$
$$ \therefore $$ Velocity in x direction,
$${v_x} = {{dx} \over {dt}} = Ky\,$$ . . . . . (1)
Velocity in y direction,
vy = $$\,{{dy} \over {dt}}$$ = Kx . . . . . . . (2)
$$ \therefore $$ $${{\,{{dy} \over {dt}}} \over {{{dx} \over {dt}}}} = {{Kx} \over {Ky}}$$
$$ \Rightarrow $$ $${{dy} \over {dx}} = {x \over y}$$
$$ \Rightarrow $$ ydy $$=$$ xdx
Integrating both sides we get,
$$\int {ydx} = \int {xdx} $$
$$ \Rightarrow $$ $${{{y^2}} \over 2} = {{{x^2}} \over 2} + c$$
$$ \Rightarrow $$ $${y^2} = {x^2} + 2c$$
$$ \therefore $$ General equation,
y2 = x2 + constant.
$$\overrightarrow v = K\left( {y\widehat i + x\widehat j} \right)$$
$$ \therefore $$ Velocity in x direction,
$${v_x} = {{dx} \over {dt}} = Ky\,$$ . . . . . (1)
Velocity in y direction,
vy = $$\,{{dy} \over {dt}}$$ = Kx . . . . . . . (2)
$$ \therefore $$ $${{\,{{dy} \over {dt}}} \over {{{dx} \over {dt}}}} = {{Kx} \over {Ky}}$$
$$ \Rightarrow $$ $${{dy} \over {dx}} = {x \over y}$$
$$ \Rightarrow $$ ydy $$=$$ xdx
Integrating both sides we get,
$$\int {ydx} = \int {xdx} $$
$$ \Rightarrow $$ $${{{y^2}} \over 2} = {{{x^2}} \over 2} + c$$
$$ \Rightarrow $$ $${y^2} = {x^2} + 2c$$
$$ \therefore $$ General equation,
y2 = x2 + constant.
Comments (0)
