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Motion in A Plane question

2010 · Shift 0 · Q77
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Motion in A Plane question

2010 · Shift 0 · Q77

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A particle is moving with velocity v→=k(yi^+xj^)\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)v=k(yi+xj​), where K is a constant. The general equation for its path is
  1. A
    y = x2 + constant
  2. B
    y2 = x + constant
  3. C
    xy = constant
  4. D
    y2 = x2 + constant
View written solutionFree

Correct answer: D

  1. The velocity of the particle is given by v⃗=k(yi^+xj^).\vec v = k(y\hat i + x\hat j).v=k(yi^+xj^​).

  2. Therefore, the components of velocity are vx=dxdt=ky,v_x = \frac{dx}{dt} = ky,vx​=dtdx​=ky, vy=dydt=kx.v_y = \frac{dy}{dt} = kx.vy​=dtdy​=kx.

  3. To find the path, eliminate time: dydx=dy/dtdx/dt=kxky=xy.\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{kx}{ky} = \frac{x}{y}.dxdy​=dx/dtdy/dt​=kykx​=yx​.

  4. Rearranging, y dy=x dx.y\,dy = x\,dx.ydy=xdx.

  5. Integrating both sides, ∫y dy=∫x dx\int y\,dy = \int x\,dx∫ydy=∫xdx y22=x22+C.\frac{y^2}{2} = \frac{x^2}{2} + C.2y2​=2x2​+C.

  6. Multiply by 2 and absorb the constant: y2=x2+C.y^2 = x^2 + C.y2=x2+C.

  7. Hence, the general equation of the path is y2=x2+constant.y^2 = x^2 + \text{constant}.y2=x2+constant.

  8. Checking options:

    • A: y=x2+constanty = x^2 + \text{constant}y=x2+constant ❌
    • B: y2=x+constanty^2 = x + \text{constant}y2=x+constant ❌
    • C: xy=constantxy = \text{constant}xy=constant ❌
    • D: y2=x2+constanty^2 = x^2 + \text{constant}y2=x2+constant ✅

Therefore, the correct option is D.

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