# Homework Help: SR Velocity Transformation

1. May 29, 2016

### rbn251

1. The problem statement, all variables and given/known data
Two rockets A and B are moving away from the earth in opposite directions at 0.85c and -0.75c respectively.
How fast does A measure B to be travelling?

Now I have worked out v = -0.85-0.75/(1- -0.85*-0.75) = -0.997. This is correct.

Now I would like to work it out backwards to check my understanding so:
According to A:
B travels at 0.997c
The earth frame travels at 0.75c

Q - How fast does the earth measure B to be travelling?

2. Relevant equations
w=u-v/(1-uv)

3. The attempt at a solution

I expected the answer to be 0.85c but:
v=0.997-0.75/(1- 0.997*0.75) == 0.979 and not 0.85.

Thanks for any help

2. May 29, 2016

### PeroK

Are you sure?

Technically you're missing some brackets in that equation.

3. May 29, 2016

### rbn251

As far are I know everything is correct 'mathematically'. You are right about the brackets but I have used the equation correctly in both cases as such.

4. May 29, 2016

### stevendaryl

Staff Emeritus
Your problem is just keeping your variables straight. In my opinion, it's always good to make up variables for all the values in a problem, and do as much of the derivation as possible symbolically, and only substitute numbers into the equations at the last step.

So, let's make up some variable names:
$v_{AE} =$ velocity of $A$ as measured in the Earth's rest frame
$v_{BE} =$ velocity of $B$ as measured in Earth's rest frame
$v_{BA} =$ velocity of $B$ as measured in A's rest frame

The velocity addition formula tells you that:

$v_{BE} = \dfrac{v_{BA} + v_{AE}}{1 + \frac{v_{BA} v_{AE}}{c^2}}$

You have: $v_{BA} = -0.997c$, $v_{AE} = 0.85c$.

: $v_{BA}$ is negative, so it should be $-0.997c$.
[second edit]: As PeroK points out, it should actually be $-0.977c$

5. May 29, 2016

### PeroK

I think it should be 0.977 not 0.997. That's the problem.

6. May 29, 2016

### rbn251

haha I think you're right - typo in the text book!

7. May 29, 2016

### rbn251

yup f*s sry for bothering!

8. May 29, 2016

### PeroK

I prefer:

$v' = \frac{u+v}{1+uv}$

$v'' = \frac{v'-u}{1-v'u} = \frac{v(1-u^2)}{1-u^2} = v$

Then you can relax!

9. May 29, 2016

### stevendaryl

Staff Emeritus
But my point is that for each of the variables $u, v, v', v''$ you need to get clear in your mind: Whose velocity is it, and relative to which frame? And you also have to keep in mind that velocity has a direction, as well as a magnitude. (In these 1-D problems, "direction" means "sign")