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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?

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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.

Frames

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 $-(x+y)$.

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  • $\begingroup$ 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? $\endgroup$ – Farcher Aug 1 at 9:58
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    $\begingroup$ 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. $\endgroup$ – John Rennie Aug 1 at 10:00
  • $\begingroup$ @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? $\endgroup$ – Atharav Karhad Aug 1 at 13:27
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    $\begingroup$ 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. $\endgroup$ – John Rennie Aug 1 at 13:58
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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.

enter image description here

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 $A$.

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enter image description here 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

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

your example

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

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

$\Rightarrow$

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

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

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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

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

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