# Motion between two particles in a relative manner

Suppose a particle A is travelling in east direction with velocity of x m/s and another particle B is travelling with velocity y m/s in the west direction. Why does the the particle B appears to move towards A with a velocity of x+y and not just y m/s?

The top diagram shows the velocities in the lab frame. Particle $$A$$ is moving east at speed $$x$$ and particle $$B$$ is moving west at speed $$y$$. I'm taking the east direction to be positive, so the velocity of $$A$$ is positive and the velocity of $$B$$ is negative.

To find the velocity of $$B$$ relative to $$A$$ we have to transform to the frame where $$A$$ is stationary, and we do this by adding the velocity $$-x$$ to everything as shown in the middle diagram. Then the velocity of particle $$A$$ is $$v_A = x + (-x) = 0$$, so this makes particle $$A$$ stationary as we require.

And as the bottom diagram shows, when we add $$-x$$ to the velocity of article B we get $$v_B = -y + (-x) = -(x + y)$$. That's why the velocity of $$B$$ relative to $$A$$ is $$-(x+y)$$.

• The label above the right hand arrow should be $y$ not $-y$ as you already included the direction and are stating what the magnitude of the vector is? – Farcher Aug 1 at 9:58
• The numbers $x$ and $y$ are speeds not velocities. I chose this representation to make it explicit that in my chosen coordinates the velocity of $b$ is negative. It's always a slightly vexed issue how best to describe this, but hopefully the diagram makes everything clear. – John Rennie Aug 1 at 10:00
• @John Rennie the transformation of frame method is self created way to explain or is it an official way of explanation.Also why do we make particle A stationary? – Atharav Karhad Aug 1 at 13:27
• Hi Atharav. Transforming into the rest frame of particle $A$ is a standard way to approach these problems, and that's why I've done it in some detail. The reason we make $A$ stationary is because we want the velocity of $B$ relative to $A$ i.e. if $A$ were stationary then how fast would $B$ be approaching it. – John Rennie Aug 1 at 13:58

Suppose body $$A$$ is going to the right with a speed $$x$$ and another body $$B$$ is going to the left at a speed $$y$$.

The motion of the bodies can be represented as vector diagram 1.

To both motions add a velocity to the left of equal magnitude to that of the velocity of body $$A$$, ie stopping body $$A$$, as shown in vector diagram 2.

On adding the two vectors body $$A$$ is at rest and body $$B$$ is moving at speed $$x+y$$ to the left as shown in vector diagram 3 and this is the velocity of body $$B$$ relative to body $$A$$.

In symbols let $$\hat l$$ and $$\hat r$$ be unit vectors in the left and right direction such that $$\hat l= - \hat r$$.

Step 1 - The velocity of body $$A$$ is $$x\hat r$$ and than of body $$B$$ is $$y\hat l$$.

Step 2 - To both motions add a velocity to the left of equal magnitude to that of the velocity of body $$A$$ $$(x\hat l)$$

Step 3 - Velocity of $$A$$ is $$x\hat r + x \hat l = x(-\hat l) + x \hat l =0$$ and velocity of $$B$$ is $$y\hat l + x \hat l = (x+y) \hat l$$ and this is the velocity of body $$B$$ relative to body $$A$$.

given two vectors $$\vec{v}_{01}$$ and $$\vec{v}_{02}$$

thus:

$$\vec{v}_{12}=\vec{v}_{10}+\vec{v}_{02}\,,\quad\text{("zero cancel")}$$

where

$$\vec{v}_{10}=-\vec{v}_{01}$$

$$\vec{v}_{01}=x$$

$$\vec{v}_{02}=-y$$

$$\Rightarrow$$

$$\vec{v}_{12}=-x-y=-(x+y)\quad \surd$$

$$\vec{v}_{21}=-\vec{v}_{12}=(x+y)$$

As it is already answer by so many but i want to add one more point ,THAT IS -change in distance of B with repect to time will be equal to ym/s if you take your frame of reference as origin according to origin it velocity changes because it s distance with respect to time change equal to y=m/s but now yiu change frame of reference that is A so the velocity will be according to the observer A hence he feel that you are going (x+y)m per second ,yiu can add sign according to wether they are approaching each other or going away from each other

• This assumes Galilean transformation instead of Lorentz. For small speed it may be reasonable, but it's not fully correct. – Kyle Kanos Aug 2 at 10:17