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QJEEM19Phcb98: JEE Main 2019 Physics - Class 11 JEEM19Phcb98 Solution: A particle is moving with a velocity where K is a constant. The …

A particle is moving with a velocity

v→ =K(yi^+xj^),\overrightarrow v \, = K(y\widehat i + x\widehat j), where K is a constant.

The general equation for its path is :

Multiple Choice Options

  • A y = x2 + constant
  • B y2 = x + constant
  • C y2 = x2 + constant
  • D xy = constant

Step-by-Step Verified Solution

Given,

v→=K(yi^+xj^)\overrightarrow v = K\left( {y\widehat i + x\widehat j} \right)

∴\therefore    Velocity in x direction,

vx=dxdt=Ky {v_x} = {{dx} \over {dt}} = Ky\,    . . . . . (1)

Velocity in y direction,

vy =  dydt\,{{dy} \over {dt}} = Kx . . . . . . . (2)

∴\therefore      dydtdxdt=KxKy{{\,{{dy} \over {dt}}} \over {{{dx} \over {dt}}}} = {{Kx} \over {Ky}}

⇒\Rightarrow     dydx=xy{{dy} \over {dx}} = {x \over y}

⇒\Rightarrow    ydy == xdx

Integrating both sides we get,

∫ydx=∫xdx\int {ydx} = \int {xdx}

⇒\Rightarrow     y22=x22+c{{{y^2}} \over 2} = {{{x^2}} \over 2} + c

⇒\Rightarrow     y2=x2+2c{y^2} = {x^2} + 2c

∴\therefore    General equation,

y2 = x2 + constant.

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