Explanation

$x = x_{0} + (ut + \frac{1}{2}a t^{2})$ At $t = 0$, $x = -2$, $\therefore -2 = x_{0} + 0$ or $x_{0} = -2$ Thus, $0 = -2 + (u \times 1 + \frac{1}{2} \times a \times 1^{2})$ and $6 = -2 + (u \times 2 + \frac{1}{2} \times a \times 2^{2})$ After solving equations, we get $u = 0$, $a = 4 m/s^{2}$. Now for $t = 3$, $x = -2 + (u \times 3 + \frac{1}{2} \times 4 \times 3^{2}) = 16 m$. Clearly it represents motion with constant acceleration.