Solving Relative Speed and Flashes in S' Frame

[Oerg]

Homework Statement

A red light flashes at position xR = 3:00m and time tR = 1e-9s, and
a blue light flashes at xB = 5:00m and tB = 9e-9 s, all measured in
the S reference frame. Reference frame S` has its origin at the same point
as S at t = t0 = 0; frame S' moves uniformly to the right. Both flashes
are observed to occur at the same place in S'. (a) Find the relative speed
between S and S`. (b) Find the location of the two flashes in frame S`.
(c) At what time does the red flash occur in the S' frame?

Homework Equations

The Attempt at a Solution

I am going crazy! This is my approach.

taking the distance of S' from x_R to be L_p measured in S and distance of S' from x_B to be contracted length L_C as measured in S',

L_p=L_c \times \gamma
x_A+2=x_A \times \gamma

at t_B distance traveled by S' frame is v X t'
taking t' to be t_B \times gamma,
x_A=3-vt'

But when i substituted x_A back into the equation, i became stuck! so obviously something is wrong but i cannot figure out what is wrong.

 

[Doc Al]

I don't quite understand what you are doing. In any case, for a problem like this, why not use the Lorentz transformations directly. That's what they are for!

 

[Oerg]

Doc AI, thanks for helping me again. I have another question though that is slightly unrelated to this.

Given that f=\frac{\bar u \bar v}{\bar u +\bar v}

show that

e_f=f^2({\frac{e_u}{\bar u^2} + \frac{e_v}{\bar v^2})

where e refers to the error. ok so I added up the fractional uncertainties and I got this

\frac{e_f}{f}=\frac{e_u}{u}+\frac{e_v}{v}+\frac{e_u+e_v}{u+v}

after some simplifying, I got to this,

e_f=f^2(\frac{e_u(u+v)}{u^2v}+\frac{e_v(u+v)}{v^2u}+\frac{e_u+e_v}{uv})

and then I realized that I could never get the answer, however, if this term was negative,
\frac{e_u+e_v}{uv}, i would get the answer perfectly, but how can it be negative? Problem is even in division, shouldn't the fractional uncertianties add up??