Answer
a) $v=0.141c$
b) $v=0.312c$
Work Step by Step
(a) We know that
$v=c\sqrt{1-(1-f)^2}$
We plug in the known values to obtain:
$v=c\sqrt{1-(1-0.0100)^2}$
$v=0.141c$
(b) We know that
$v=c\sqrt{1-(1-f)^2}$
We plug in the known values to obtain:
$v=c\sqrt{1-(1-Pl0.0500)^2}$
$v=0.312c$
