Explain the relative velocity and its two cases.

Explanation

Case 1 Two observes and one moving object :

Let A be the frame of reference associated with person standing on the ground.
B is the frame of reference associated with the train moving with uniform velocity.
Both these reference frames are inertial frames of reference.
From the figure, `x_(PA)=x_(PB) +x_(BA)`
Differentiating with respect to t,
`(dx _(PA))/(dt) = (dx _(PB))/(dt )+ (dx _(BA))/(dt)`
But ` (dx _(PA))/(dt) = v _(PA),( dx _(PB))/(dt)= v _(BA)`
`therefore v (PA) = v _(PB) + v _(BA)`
`therefore v _(BA) = v _(PA) - v_({PB)`
Where `v _(PA)=` Velocity of P with respect to reference frame A.
`v _(PB)=` Velocity of P with respect ot reference frame B.
`v _(BA)=` Velocity of reference frame B with respect to reference frame A.
Case 2 One observer and two moving objects :

Suppose particales A and B motion in reference of frame O- XY O - YZ in ground.
Suppose velocities of two particles A and B are respectively `vec(v _(A)) and vec(v_(B))` releative to a farme of reference (suppose earth) then velocity `v _(AG) and v _(GB).` Then velocity of B relative to A is,
`v _(BA)= v _(GA) - v _(GB)`
`=- v _(GB) + v _(GB)`
` therefore v_(BA) =-v _(BG)-v_(AG)`
`[because v_(BG) =- v _(GB) and v _(GA) =-v _(AG) ]`
`=v _(B) -v_(A)`
and velocity of A relative to B is
`v_(AB) =v_(A) -v_(B)`
`= v _(AG)-v_(BG)`