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Question

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The spaceship Enterprise 1 is

moving directly away from earth at a velocity that an earth-based

observer measures to be +0.75c. A sister ship, Enterprise 2, is

ahead of Enterprise 1 and is also moving directly away from earth

along the same line. The velocity of Enterprise 2 relative to

Enterprise 1 is +0.42c. What is the velocity of Enterprise 2, as

measured by the earth-based observer?

Â

I have been getting "sum the two togeather" but somewhere the

formula Vab=(Vac+Ccb)/(1+(VacVcb/c^2)) has to be used. the

instructor wrote =0.86c on the test question I missed. I dont know

how he got there. I am confused.

or maybe the relativity Change T=T/sqrt 1-(v/c)^2 in relation to ship 1 needs to be used

Explanation / Answer

the formula to be used is

v' = (V1+V2)/(1+(V1*V2/C^2))

thus is the formula to be used for velocities close to that of light. the velocity cannot be more than that of light. due to contranction of length (relativistic length contraction)

thus

V' = (0.75c +0.42c)/(1+(0.75c +0.42c)/c^2) = (0.75 +0.42)/(1+(0.75 *0.42) = 0.889733840304 c

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