Lorentz velocity transformation problem

[Aziza]

An observer on Earth observes two spacecraft moving
in the same direction toward the Earth. Spacecraft A appears
to have a speed of 0.50c, and spacecraft B appears
to have a speed of 0.80c. What is the speed of spacecraft
A measured by an observer in spacecraft B?So if S is the reference frame of the Earth, S' is the reference frame of plane B, v is the velocity with which S' approaches S (according to S), u is the velocity with which plane A approaches Earth (according to Earth) and u' is the velocity with which plane A approaches plane B (according to B), then:

v = -0.8c
u = -0.5c

u' = (u-v) / 1-(uv/c²)) = 0.5c

I am interpreting this as meaning that according to plane B, plane A is actually moving away from B at 0.5c. My book, however, has the answer as -0.5c, implying the opposite, i think?...is the book wrong or are my signs wrong? It makes sense to me making v and u negative because the formula for the velocity transformation was derived assuming that S' and the event (plane A here) are moving away from S, not towards...

Homework Statement

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

Hi Aziza!

I think you're over-thinking this …

An observer on Earth observes two spacecraft moving
in the same direction toward the Earth. Spacecraft A appears
to have a speed of 0.50c, and spacecraft B appears
to have a speed of 0.80c. What is the speed of spacecraft
A measured by an observer in spacecraft B?

… if we ignore relativity, and choose 50mph and 80mph instead,

then obviously B is going faster than A, so A's speed measured by B will be negative

(of course, this is a non-standard usage of "speed" … the standard usage is as the magnitude of velocity, so that it's always non-negative)