Unable to clearly understand relative velocity

[JC2000]

Unable to understand why this is the case?
Also, if it were in two dimensions then I am guessing the same would apply, only the 'x-co-ordinates' and 'y-coordinates' would add up separately?
Lastly, does the same hold for acceleration.

I am aware that this is a trivial question, many books address it but I am unable to 'explain'/understand this...
Thank you!

 

[Herman Trivilino]

Summary: If particle $A$ were to be moving with speed $v$ (with reference to a fixed point) towards a particle $B$ which is moving towards $A$ with speed $u$ (with reference to the same fixed point), then why is the relative velocity of $A$ with respect to $B$ be $v+u$?

Unable to understand why this is the case?

Can you elaborate more on why you're unable to understand? If your fixed point is at rest relative to, say, a roadway, and two cars approach each other on that roadway, wouldn't you expect to add the speeds?

 

[Gaussian97]

Hi, Suppose that you have two reference frames, the Reference Frame RF1 is at the origin (according to RF1) while the second Reference Frame RF2 is at some position $x$ (let's do it along one axis). Imagine that this RF2 is moving with respect to RF1 with velocity $u$, which means that, after some time $\Delta t$ the reference frame RF2 will be in another position $x+u\Delta t$. Now imagine some object at a distance $r$ from the RF1 that is also moving with velocity $v$, which means that after some time $\Delta t$ the object will be at a distance $r+v\Delta t$.
First of all, in the beginning, when RF2 is at a position $x$ and the object is at $r$, what is the distance between RF2 and the object? (Let's suppose that both quantities are positive and $r>x$.) Well, then I think you will agree that the distance between them is exactly $$r'=r-x$$ (if you don't understand why, try to draw it.)
So, as I said, after a time $\Delta t$ the reference frame is at a new position $x+u\Delta t$, while the object is at $r+v\Delta t$. For the same argument, what is now the distance between them? Again is $$\left(r+v\Delta t\right)-\left(x+u\Delta t\right)=\left(r-x\right)+\left(v-u\right)\Delta t=r'+\left(v-u\right)\Delta t$$. What is the velocity that will see the RF2? Well, after a time $\Delta t$ the object has move from $r'$ to $r'+\left(v-u\right)\Delta t$, so we can compute the velocity in the following way: $$v'=\frac{\Delta r'}{\Delta t}=\frac{r'+\left(v-u\right)\Delta t-r'}{\Delta t}=v-u$$. So the velocity of the object with respect to RF2 is exactly $v'=v-u$.

In this case, I have assumed that both objects have positive velocities and positive positions (although is not necessary, these equations are still true otherwise). So what happens when the object has negative velocity (i.e. it goes towards RF2) then the equation is exactly the same $v'=v-u$ but now $u<0$. Since we usually prefer positive numbers, we can change the equation to include the $-$ sign, then: $$v'=v+u$$ where now both $v$ and $u$ are positive numbers.