Velocity of Separation in Special Relativity

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An observer sees two spaceships moving apart at a speed of 0.99c. The relative velocity between the two spaceships can be calculated using the formula v_rel = (v_A + v_B) / (1 + (v_Av_B/c^2)). However, without knowing the individual velocities vA and vB, the question lacks sufficient information to determine the relative speed. Clarification suggests that both ships are traveling at 0.99c relative to a stationary observer in opposite directions. The correct relative speed, as confirmed by a reference book, is approximately 0.99995c.
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An observer sees two spaceships flying apart with speed .99c. What is the speed of one spaceship as viewed by the other?

I considered two spaceships A and B. A is moving with velocity vA in the left direction and B is moving with velocity vB in the right direction. Hence, ##v_{sep} = v_A + v_B = 0.99c##

Now finding the relative velocity of one spaceship with respect to other.

$$v_{rel} = \frac{v_A + v_B}{1 + \frac{v_Av_B}{c^2}}$$

Now since I don't know vA or vB, I can't find the answer. So, my question here is - is my question incomplete? Or is it possible to obtain an answer.
 
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Your question is incomplete. you have the sum vA+vB but not the product. The product of two numbers cannot be obtained from their sum without some additional information.
 
Maybe the question means to say each ship is traveling at 0.99c relative to the stationary observer in opposite directions.
 
vela said:
Maybe the question means to say each ship is traveling at 0.99c relative to the stationary observer in opposite directions.

Hmm... I tried to look for the answer in the meantime, the book (Kleppner and Kolenkow) from which the question was lifted, fortunately provided it.

Taking your suggestion, the answer is coming out to be 0.99995c, which is the same as the answer provided.

Thanks a ton.
 

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