Solution 7.3a
From Mechanics
\displaystyle t=0\ \ gives
\displaystyle \mathbf{r}=\left( 2\times 0-16 \right)\mathbf{i}+\left( 4\times 0-12 \right)\mathbf{j}=-16\mathbf{i}-12\mathbf{j}\ \text{m}
\displaystyle t=40\ \ gives
\displaystyle \mathbf{r}=\left( 2\times 40-16 \right)\mathbf{i}+\left( 4\times 40-12 \right)\mathbf{j}=64\mathbf{i}+144\mathbf{j}\ \text{m}
\displaystyle t=80\ \ gives
\displaystyle \mathbf{r}=\left( 2\times 80-16 \right)\mathbf{i}+\left( 4\times 80-12 \right)\mathbf{j}=144\mathbf{i}+308\mathbf{j}\ \text{m}
\displaystyle t=120\ \ gives
\displaystyle \mathbf{r}=\left( 2\times 120-16 \right)\mathbf{i}+\left( 4\times 120-12 \right)\mathbf{j}=224\mathbf{i}+468\mathbf{j}\ \text{m}