Question

A particle moves in x-y plane with a velocity v = 2y i +4j. Equation of the path followed bythe particle is
(1) x = √y
(2) y = √x
(3) 4x = y2
(4) 4y = x^2​

Answer 1

Answer:

(3) 4x = y^2

Explanation:

Given the velocity vector v = 2yi + 4j

Horizontal component of the velocity = 2y

Vertical component of the velocity = 4

We know that velocity is nothing but rate of change of displacement

Therefore, in the x-direction,

$\frac{dx}{dt} =2y$

And, in the y-direction,

$\frac{dy}{dt} =4$$\frac{dy}{dx} =\frac{2}{y}$

Since, $\frac{dy}{dt} /\frac{dx}{dt} =\frac{dy}{dx}$

Therefore,

$\frac{dy}{dx} =\frac{4}{2y}$

or, $\frac{dy}{dx} =\frac{2}{y}$

or, $ydy=2dx$

or, ∫ydy = 2∫dx

or, $\frac{y^{2} }{2} =2x+c$ (where c is a constant)

Assuming Particle to be at rest initially we get c=0

Thus we get the path equation of the particle

$y^{2} =4x$