Answer
$31mi/h$
Work Step by Step
The required speed can be determined as follows:
$v=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
We plug in the known values to obtain:
$v=\frac{22mi/h+19mi/h}{1+\frac{(22mi/h)(19mi/h)}{(35mi/h)^2}}$
$v=31mi/h$
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: 39
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