Knowing that acceleration is time derivative of velocity, and (if you need) velocity is time derivative of position
$v = \dot x$
$a = \dot v = \ddot a$
the equation of your question can be written as a simple first-order linear ordinary differential equation,
$\dot v + k v = g$
whose solution reads
$v(t) = \dfrac{g}{k} + A e^{-kt}$.
Prodivded the initial condition $v(0) = v_0$, it's possible to determine the constant $A$,
$A = v_0 - \dfrac{g}{k}$
to get the solution of the problem as
$v(t) = \dfrac{g}{k} \left( 1 - e^{-kt}\right) + v_0 e^{-kt} $.