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Motion in A Plane question

2019 · 9 Jan · Shift 1 · Q52
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Motion in A Plane question

2019 · 9 Jan · Shift 1 · Q52

JEE MainPhysicsMotion in A PlaneMCQ+4 / −1
A particle is moving with a velocity v→ =K(yi^+xj^),\overrightarrow v \, = K(y\widehat i + x\widehat j),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
    y2 = x2 + constant
  4. D
    xy = constant
View written solutionFree

Correct answer: C

  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. Compare with the general form v⃗=vxi^+vyj^,\vec v = v_x \hat i + v_y \hat j,v=vx​i^+vy​j^​, so we get 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. Integrate 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: y2=x2+C′,y^2 = x^2 + C',y2=x2+C′, where C′C'C′ is another constant.

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

  8. Check options:

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

Therefore, the correct option is C.

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