Answer:
42.11 years old
Explanation:
Given that:
In 2000, a 20-year-old astronaut left Earth to explore the galaxy; her spaceship travels at 2.5 x 10^8 m/s. She returns in 2040
To find her age we use:
[tex]\Delta t_m=\frac{\Delta t_s}{\sqrt{1-\frac{v^2}{c^2} } }\\[/tex]
Δtm is time interval for the observer stationary relative to the sequence of
events = 2040 - 2000 = 40 years
Δts is is the time interval for an observer moving with a speed v relative to the sequence of event
v = velocity = 2.5 x 10^8 m/s
c = speed of light = 3 x 10^8 m/s
[tex]\Delta t_s=\Delta t_m}{\sqrt{1-\frac{v^2}{c^2} } }\\\Delta t_s=40\sqrt{1-\frac{(2.5*10^8)^2}{(3*10^8)^2}}\\\Delta t_s=22.11\ yr[/tex]
Here age in 2000 is 20 year, therefore when she appear she would be 20 year + 22.11 year = 42.11 years old
