Question

a partical moves in x-y plane with a velocity v=2yi^+4j^ equation of the path followed by the particle is

Answer 1

answer : option (3) y² = 4x

explanation : A particle moves in x-y plane with a velocity , v = 2y i + 4 j.

we know, rat of change of displacement with respect to time, is known as velocity.

so, v = (dx/dt)i + (dy/dt)j = 2y i + 4j

on comparing we get,

dx/dt = 2y and dy/dt = 4

from dy/dt = 4

or, ∫ dy = 4∫dt

or, y = 4t .....(1)

from dx/dt = 2y

or, dx/dt = 2(4t) = 8t [ from equation (1), ]

or, ∫dx = 8∫t.dt

or, x = 8 × t²/2

or, x = 4t² ......(2)

from equations (1) and (2),

x = y²/4 => y² = 4x

hence, equation of path followed by the particle is 4x = y²

Answer 2

Answer:

option 3:

Explanation:

see the attached image