# Perception of faster than light travel?

Tags:
1. Oct 19, 2014

### metal_maniac

I'm currently taking a modern physics course, I came across this problem which really threw me off guard:

Three spaceships A, B, and C are in motion as shown in the figure. The commander on ship B observes ship C approaching with a relative velocity of 0.83c. The commander also observes ship A, advancing in the rear, with a relative velocity of 0.48c. As measured by commander on ship B, at what speed is ship A approaching ship C?

Attempt at a solution: I used the transformations equations which resulted in a value of 0.94c.
observed velocity = [(-0.83c) - 0.48c]/[1-(0.83c*0.48c/c^2)] = 0.937c
0.94c was one of the possible answers so I picked it... it was wrong.
The correct answer was 1.3c.

Is this even possible?
I do suppose that someone with no knowledge in physics would just add the two velocities like it was Galilean transformations, but we're taking about the commander of a ship, pretty sure he wouldn't just add them together.
Is this just a badly written question or am I missing some things in my understanding? Any help is appreciated!
Thank you

2. Oct 19, 2014

### ShayanJ

The relativistic formula for addition of velocities is used when you want to change frames.
Here things are different. You're actually calculating the closing speed of spaceships A and C in the B's frame. He sees A to travel at 0.48c and C to travel at 0.83c, so if he attempts to find the time derivative of the distance between the two, he has no way other than simply adding their velocities.
I should say that this doesn't violate SR since there is no mass or energy going faster than light!
Take a look at here for more information!

Last edited: Oct 19, 2014
3. Oct 19, 2014

### metal_maniac

You're a life saver!
I was stuck on this question for so long, and my textbook had no explanation for it.
Thank you so much!