# A Roller Coaster Ride through Relativity/Appendix D

## The Addition of SpeedsEdit

Consider a spaceship A of length *l*_{A} travelling past you at a speed *v*_{A} passing a second spaceship B of length *l*_{B} travelling in the opposite direction at a speed *v*_{B} (both speeds measured with reference to you). The question we need to answer is what speed *v*_{X} does B appear to be going past the astronauts in A?

We need to consider two vital events; first contact, when the nose cones of the two spaceships meet and second contact when the two tails part.

What is more, let us suppose that these two events occur *at the same place* from your point of view. This means that the spaceships have to be just the right length so that they both pass you in the same time. Note that from your point of view, both ships are contracted by the factors *γ*_{A} and *γ*_{B} respectively.

Now what does the commander of ship A see? He sees *you* travelling past at a speed *v*_{A} and also ship B travelling past at some greater speed *v*_{X}.

Note that to the occupants of ship A, B is length contracted by a factor *γ*_{X}.

Now commander A can calculate the time between first and second contact in two ways. First he sees you travelling a distance *l*_{A} at a speed *v*_{A}. Secondly, he sees ship B (whose length is contracted to *l*_{B}/*γ*_{X}) travel a distance equal to *l*_{A} + *l*_{B}/*γ*_{X} at a speed *v*_{X}. It follows that:

An identical argument from commander B's point of view leads to:

(It should be pointed out that because of the invariance of velocity, if commander A sees ship B moving at a speed *v*_{X} then commander B will see ship A moving at exactly the same speed and contracted by the same γ factor.)

Now multiplying the two equations together eliminates the lengths of the two ships and leaves us with a relation between the three velocities. The rest is just algebra.

Now

so:

hence:

from which we obtain by straightforward algebra the result we desire: