Relative Velocities: Speed of C in A's Frame

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In the problem involving trains A, B, and C, the length of train B in A's frame is 4L/5, while the lengths of A and B are equal in C's frame. The solution involves using the Lorentz contraction formula to find the speed of C in A's frame, leading to the equation u = 3/5. The relationship between the relative speeds is established using the velocity addition formula, resulting in a quadratic equation for v. The discussion indicates that the next step is to solve this quadratic equation for v, confirming the calculations are on the right track. The focus remains on determining the correct speed of train C relative to A.
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Homework Statement


Three trains A, B, and C with equal proper lengths L are moving on parallel tracks. In the frame of A, B has length 4L/5. In the frame of A, what is the speed of C, if the lengths of A and B are equal in the frame of C

Homework Equations


u'=[u-v]/[(1-uv)]
c=1

The Attempt at a Solution



L' = L/γ => 4L/5 = L*sqrt(1-u^2) => sqrt(1-(4/5)^2) = u => sqrt[(25-16)/25] = u => u= 3/5

Where:
u'=v since is the relative speed of C/B or C/A since B/C=A/C. This is true because L is the same for all trains, and both trains A and B have the same length in C?
v = relative speed of C/A
u = relative speed of B/A

Thus
v=[u-v]/[(1-uv)] => v*(1-uv)=[u-v] => v-v^2*u-u+v=0 => -v^2+(2v/u)-1 = 0 => v^2-(10v/3) +1 = 0
 
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messier992 said:
v=[u-v]/[(1-uv)] => v*(1-uv)=[u-v] => v-v^2*u-u+v=0 => -v^2+(2v/u)-1 = 0 => v^2-(10v/3) +1 = 0
I think this is OK. Did you go on to solve the last equation for v?
 
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