Question:

A particle is moving with velocity $\vec{v} = k\left(y\hat{i} + x\hat{j}\right),$ where $k$ is a constant. The general equation for its path is

Updated On: Jul 2, 2022
  • $y = x^2 +$ constant
  • $y^2 = x +$ constant
  • $xy =$ constant
  • $y^2 = x^2 +$ constant
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The Correct Option is D

Solution and Explanation

$\vec{v} = K\,y\hat{i} + K\,x\hat{j}$ $\frac{dx}{dt} = Ky, \quad \frac{dy}{dt} = Kx$ $\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{Kx}{Ky}$ $y \,dy = x\, dx$ $y^{2} = x^{2} + c.$
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Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration