Answer
$0.98c$
Work Step by Step
We know that
$v_{12}=\frac{v_{1E}+v_{E2}}{1+\frac{v_{1E}v_{E2}}{c^2}}$
We plug in the known values to obtain:
$v_{12}=\frac{0.8c+0.8c}{1+\frac{(0.8c)(0.8c)}{c^2}}$
$v_{12}=\frac{1.6c}{1+0.64}$
$v_{12}=0.98c$
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: 41
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