You can't go faster than light, and light can't be additive (if you shine a light from a spaceship, the light is not going $c$+"speed of spaceship", it's just going like it always does).
But what happens when two spaceships, each going 75% the speed of light, are traveling directly towards each other?
Aren't the two spaceships now traveling, in reference to each other, at 150% the speed of light?
The addition of relative velocities is $$\frac{a+b}{1+ab}$$ so $\frac{.75+.75}{1+.{75}^2}$ = .96 c.
Consider that you are travelling almost the speed of light (1-x) c and you see a space ship pass you at an equal speed. Combining (1-x) c with (1-x)c. This gives $$\frac{2-2 x}{2-2 x+x^2}$$ Since $2-2 x < 2-2 x+x^2$, we have $\frac{2-2 x}{2-2 x+x^2} <1$
The addition of relative velocities is bound by the mathematics of the hyperbolic tangent. So velocity $a$ plus velocity $b$ is the inverse hyperbolic tangent of $a$ plus the inverse hyperbolic tangent of $b$ and then finally take the hyperbolic tangent of the results. The hyperbolic tangent only goes to 1 at the limit of infinity.
The Special Theory of Relativity deals with two observers, well... observing a single event, and comparing their observations. So:
You observe two spaceships, S-A and S-B, occupied by Obs-A and Obs-B moving towards you at $0.75c$ from opposite directions.
You observe S-A travelling towards you at $0.75c$; Obs-A observes you moving towards him at $0.75c$. Nothing strange here.
You turn in the opposite direction and observe S-B travelling towards you at $0.75c$; Obs-B observes you moving towards him at $0.75c$. Nothing strange here, either.
You conclude, correctly, that S-A and S-B are approaching each other at $1.5c$. If they started out 30 light-years apart, they'll meet in 20 years, right about where you're standing.
But..
Obs-A will measure Obs-B to be approaching Obs-A at the velocity given in the answer above: $0.96c$. Note: MEASURE
The SPoR lets everyone in these three frames of reference predict what everyone else will observe, what their clocks will read at various events, and why various slow clocks and shortened meter-sticks make everything come out right in the end.
No, unfortunately this is one of those "It should be, but it isn't."
Ship A shines a light at ship B. The light leaves ship A at $c$ and it arrives at ship B at $c$. Even if the 2 ships are traveling at 99% $c$, the light still leaves ship A at $c$ and arrives at ship B at $c$. I believe that the reason lies with the time dilation when approaching $c$.
Alice places herself at system C. Now she detect A and B approaching each other from opposite directions, while A and B both have velocity 0.9c. She can safely say, B is approaching A with velocity 1.8c. However, if Bob is located at A and detect the velocity B towards him, he will find that B is approaching with velocity ~0.994c. If Alice and Bob can communicate with no latency(of course, this is against Einstein's theory, since communicating speed cannot be faster than c):
Alice: B is coming to you at velocity 1.8c, now you and B has distance 1.
Bob: No, B is coming to me at velocity 0.994c, and the distance from B to me is only 0.436!
It only depends where you are doing experiments, C or A.
