Answer:
The star will have a transverse speed of 315950.9 AU/year
Step-by-step explanation:
d = 1/p
d = distance to star, measured in parsecs
p = parallax, measured in arcseconds = 0.5 arcsec/year
So, d = 1/0.5 = 2 parce
1 parsec = 3.26 light years
2 parce = 6.52 light years
⇒Transverse speed in AU / year = Distance/parallax
distance = 10pc = 2060000 AU
Transverse speed in AU / year = 2060000 Au/6.52 light years
Transverse speed = 315950.9 AU/year
Therefore, A star (not Barnard's star) at a distance of 10 pc observed to have a proper motion of 0.5 arcsec / year. Will have a transverse speed of 315950.9 AU/year