From Mechanics

We use the result from part a).

\displaystyle \mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{k}\ , for any time \displaystyle t.

At the end when \displaystyle t=40 also from part a), we get

\displaystyle \mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j}) \times 40+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}) \times{{40}^{\ 2}}+10\mathbf{k}=328\mathbf{i}+336\mathbf{j}+10\mathbf{k} \ , for any time \displaystyle t.

The car starts at the origin and finishes at \displaystyle 328\mathbf{i}+336\mathbf{j}.

The magnitude of this position vector is the distance \displaystyle D\ travelled.

\displaystyle D=\sqrt{{{328}^{2}}+{{336}^{2}}}=\sqrt{107584+112896}=\sqrt{220480}=470\ \text{m}