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	<math>\sqrt{{{7}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{13}=3\textrm{.}6\ \text{m}</math>		<math>\sqrt{{{7}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{13}=3\textrm{.}6\ \text{m}</math>
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		+	In the second part the boat travel with constant velocity <math>6\mathbf{i}-8\mathbf{j}</math>
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	Thus during the first part the boat has travelled a distance 3.6 m.		Thus during the first part the boat has travelled a distance 3.6 m.

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Revision as of 19:57, 14 April 2010

Here we use

\displaystyle \mathbf{r}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t+{{\mathbf{r}}_{0}}

in the first part.

According to the text

\displaystyle \mathbf{u}=\mathbf{i}+2\mathbf{j}

\displaystyle \mathbf{v}=6\mathbf{i}-8\mathbf{j}

We assume the starting point is the origin so that \displaystyle {{\mathbf{r}}_{0}}=0.

At \displaystyle t=10

\displaystyle \mathbf{r}=\frac{1}{2}(\mathbf{i}+2\mathbf{j}+6\mathbf{i}-8\mathbf{j})t=(3\textrm{.}5\mathbf{i}-3\mathbf{j}) \times 10=35\mathbf{i}-30\mathbf{j}=(7\mathbf{i}-6\mathbf{j}) \times 5

The distance is the magnitude of this vector

\displaystyle \sqrt{{{7}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{13}=3\textrm{.}6\ \text{m}

In the second part the boat travel with constant velocity \displaystyle 6\mathbf{i}-8\mathbf{j}

Thus during the first part the boat has travelled a distance 3.6 m.