A particle moves along a horizontal straight line with an acceleration $=6 ^{1/3} m/s^{2}$. When $t = 2 s$, its displacement $S = 27 m$ and its velocity $v = 27 m/s$. Calculate the velocity and acceleration of the point when t = 4 s.

My first guess was to integrate with respect to time in order to get the velocity equation. However, this yields an equation that requires to know the position. If I integrate again, the position equation will also require position/displacement. Thus I am achieving nothing by doing so.

My professor gave us the following (among other equations):

$a=f(s)$

Hence, $a ds = v dv$

=> $v dv= f(s) dv$

How can I approach this problem? I know I have to get time as a variable somehow, but I do not know how.

Edit: I do not know if I have to use the previous identity. It's simply my next guess.