$\therefore \;v = \frac{{dy}}{{dt}} = b + 2ct - 4d{t^3}$ and $a = \frac{{dv}}{{dt}} = 2c - 12d{t^2}$
Hence, at $t = 0$, $v_{initial} \,= b$ and $a_{initial}$ $= 2c$.
$\begin{array}{|c|c|c|c|c|} \hline t( s ) & 0 & 1 & 2 & 3 \\ \hline x ( m ) & -2 & 0 & 6 & 16 \\ \hline \end{array} $