Answer
(a) The person on the airliner has aged less than friends at home.
(b) The person on the airliner aged $~~13.8~ns~~$ less than friends at home.
Work Step by Step
(a) The person on the airliner was moving at a high speed so the person on the airliner has aged less than friends at home.
(b) We can find the time of a one-way flight according to observers on the ground:
$t' = \frac{d}{v} = \frac{5.0\times 10^6~m}{250~m/s} = 2.0\times 10^4~s$
We can find the elapsed time $t_0$ in the airliner reference frame:
$t' = \gamma~t_0$
$t_0 = \frac{t'}{\gamma}$
$t_0 = t'~\sqrt{1-\frac{v^2}{c^2}}$
$t_0 = (2.0\times 10^4~s)~\sqrt{1-\frac{(250~m/s)^2}{(3.0\times 10^8~m/s)^2}}$
$t_0 = 19,999.999999993~s$
We can find the time difference:
$\Delta t = t' - t_0$
$\Delta t = (20,000~s) - (19,999.999999993~s)$
$\Delta t = 6.9~ns$
Since the flight was a return flight, the total time difference is $13.8~ns$
The person on the airliner aged $~~13.8~ns~~$ less than friends at home.
