Solution 18.6a

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\displaystyle \begin{align} & \mathbf{r}(10)=(12-2\times {{10}^{2}})\mathbf{i}+(10-5\times {{10}^{2}})\mathbf{j} \\ & =(-188\mathbf{i}-490\mathbf{j})\ \text{m} \end{align}