How Fast is Spaceship A Traveling Relative to Spaceship B?

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SUMMARY

The discussion centers on calculating the relative velocity of Spaceship A, traveling at 0.500c, as observed from Spaceship B, which is traveling at 0.800c. The correct formula for the velocity transformation is applied, resulting in the conclusion that Spaceship A is observed to be traveling at -0.500c from the perspective of Spaceship B. The error identified in the initial calculation was the incorrect inclusion of a square root in the velocity transformation formula. The final answer aligns with the textbook's solution.

PREREQUISITES
  • Understanding of special relativity concepts
  • Familiarity with the velocity transformation formula
  • Basic knowledge of the speed of light (c)
  • Ability to perform algebraic manipulations involving square roots
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  • Study the derivation of the velocity transformation formula in special relativity
  • Explore examples of relative velocity calculations in different inertial frames
  • Learn about Lorentz transformations and their applications
  • Investigate the implications of relativistic speeds on time dilation and length contraction
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Students of physics, particularly those studying special relativity, educators teaching advanced physics concepts, and anyone interested in understanding relativistic motion and velocity transformations.

Gott_ist_tot
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I am doing an extra problem from my textbook as review and I am stumped.

An Earth based observer sees to spaceships approaching Earth in the same direction. Spaceship A is traveling at 0.500c and Spaceship B is traveling at 0.800c. How fast is spaceship A traveleing as viewed by an observer on spaceship B?

I used a velocity transformation like:

(0.500c - 0.800c)/sqrt(1-[(0.5c)(0.8c)/(c^2)])

The book gets -0.500c.

I am finding it difficult to see that the Earth based observer and the spaceship moving at 0.800c observe that spaceship going the same speed.

Thanks for your help.
 
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Wow! It's always the simplest of mistakes. Thanks a lot.
 

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