Anna and Bob have identical meter sticks. Anna an observer on Earth, sees Bob traveling in a spaceship at 0.5c away from her. Bob leaves his spaceship in an escape pod, moving away from Earth and the spaceship at 0.1c with respect to the spaceship. How long is Bob's meter stick according to Anna?
x' = gamma*x
x' = gamma*(x-vt)
u = (u' + v)/(1+u'v/c2)
The Attempt at a Solution
So first I used the velocity addition equation to find the velocity of the escape pod with respect to Anna (treating the velocity of the escape pod as u' and the velocity of the spaceship with respect to Anna as v). With this, I found that the velocity of the pod relative to Earth was u=0.57c. I then used the length contraction equation (since the meter stick is proper length in the frame of Bob's shuttle), using u as my new v. This got me one answer, 0.821m.
Then, to check my answer I decided to use length contraction and time dilation to find the answer. First I converted the length of the meter stick from the pods frame to the shuttles and found it to be 0.995m. I then used this to find the time. I was a little iffy on whether I could still do the simplified length contraction equation for my second transformation, so I actually tried using both and neither "worked" with the above answer. I then tried using the time I got (cant find the numbers in my mess of work but it was ~3*10^-10 I think) and the 0.995m in the full lorentz transformation. This got me a different answer, a good .02-3 off. I also tried using just the x'=gamma*x again with similar issues.
I kind of lost my work for the end of the second method, but which ever way I chose for the second method, my answer never seemed to match my first. Did I make a math error in the first approach? Or is one of the approaches invalid?