a cyclist bikes at a speed of 6 m/s after rolling down a hill in 3 s. Of the cyclist has a constant acceleration of 2 m/s during the descent, what was the velocity of the cyclist at the top of the hill?

Answer 1

since a = 2m/s^2 , note you have a typo there.
v = 2t + c
when t = 3, v = 6
6 = 2(3) + c
c = 0

v = 2t
so at top of hill, t = 0
then v = 0