Answer
$0.38c$
Work Step by Step
We know that
$v_{12}=\frac{v_{1E}-v_{2E}}{1-\frac{v_{1E}v_{2E}}{c^2}}$
We plug in the known values to obtain:
$v_{12}=\frac{0.80c-0.60c}{1-\frac{(0.80c)(0.60c)}{c^2}}$
$v_{12}=\frac{0.20c}{1-0.48}$
$v_{12}=0.38c$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 29 - Relativity - Problems and Conceptual Exercises - Page 1042: 40
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