From Mechanics

We must find the position of the particle after 30 s.

Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} with \displaystyle \ \mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}} which was calculated in part a), \displaystyle \ \mathbf{u}=4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}} and \displaystyle {{\ \mathbf{r}}_{0}}=0 gives,

\displaystyle \mathbf{r}=(4\mathbf{i}+6\mathbf{j}) \times 30+\frac{1}{2}(\mathbf{i}+2\mathbf{j}) \times {{30}^{\ 2}}

\displaystyle \mathbf{r}=120\mathbf{i}+180\mathbf{j} +450\mathbf{i}+900\mathbf{j} = 570\mathbf{i}+1080\mathbf{j}

As the starting point is at the origin, the distance is the magnitude of this position vector.

\displaystyle \sqrt{{{570}^{2}}+{{108}^{2}}}=\sqrt{324900+1166400}=\sqrt{1491300}=1221\ \text{m}