Solution 8.6b
From Mechanics
Here we use
\displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}
According to the text
\displaystyle \mathbf{u}=\mathbf{i}+2\mathbf{j}
and part a) gave
\displaystyle \mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}
We assume the starting point is the origin so that \displaystyle {{\mathbf{r}}_{0}}=0.
At \displaystyle t=10+40=50 we obtain
\displaystyle \mathbf{r}=(\mathbf{i}+2\mathbf{j}) \times 50+\frac{1}{2}(\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j} )\times{{50}^{\ 2}}
The distance is the magnitude of this vector