Let's examine the chase scene. The morons rob a bank, and speed away at 50 mph on route 66. The police get a call of a sighting 1 mile away. They immediately speed after them at 60 mph. How long will it take to catch up with the robbers?
distance between cars = 1, robber car speed r = 50, cop car speed c = 60, d = distance.
For robbers, d=1 + rt.
For cops, d=ct.
If cops and robbers are coincident, then 1+rt = ct.
Rearranging, t(c-r)=1, and t=1/(c-r).
Note, nothing is moving at a speed of (c-r)!
There are two objects in motion, so both speeds are required to solve the problem.
Transfer this relationship to an SR example.
A stick 1 unit long with a light at the near end and a mirror at the far end, moves at v along x. A photon moves at c toward the mirror. Using the same setup for the 1st example, t1=1/(c-v). For the reflection, t2 = 1/(c+v).
Nothing moves at (c-v) or (c+v).
Those two values are referred to as 'closing speeds'. How long to form or close a spatial gap/interval.