Solution 8.6b

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Here we use
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We assume the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math>
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<math>\mathbf{r}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t+{{\mathbf{r}}_{0}}</math>
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The boat first accelerates to a point <math>A</math> say. We first must calculate the position of this point <math>{{\mathbf{r}}_{A}}</math>.
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in the first part.
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Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with,
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According to the text
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<math>t=10</math>, <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> and from part a)
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<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>
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<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>
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<math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0=35\mathbf{i}-30\mathbf{j}</math>
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<math>\mathbf{v}=6\mathbf{i}-8\mathbf{j}</math>
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The next stage has
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We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.
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<math>t=40</math>, <math>\mathbf{a}=0</math>, <math>{{\mathbf{r}}_{0}}=35\mathbf{i}-30\mathbf{j}</math>, and <math>\mathbf{u}=6\mathbf{i}-8\mathbf{j}</math> as the final position and velocity of the first stage is the initial position and velocity of the second stage.
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At <math>t=10</math>
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Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get
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<math> \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</math>
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<math>\mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+0+(35\mathbf{i}-30\mathbf{j})=275\mathbf{i}-350\mathbf{j}</math>
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The distance is the magnitude of this vector
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This is the boat´s final position.
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<math>\sqrt{{{7}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{13}=3\textrm{.}6\ \text{m}</math>
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The distance from the starting point is the magnitude of this vector,
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Thus during the first part the boat has travelled a distance 3.6 m.
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<math>\sqrt{{{275}^{2}}+{{\left( -355 \right)}^{2}}}=445\ \text{m}</math>

Current revision

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

The boat first accelerates to a point \displaystyle A say. We first must calculate the position of this point \displaystyle {{\mathbf{r}}_{A}}.

Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} with,

\displaystyle t=10, \displaystyle \mathbf{u}=\mathbf{i}+2\mathbf{j} and from part a) \displaystyle \mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}

\displaystyle {{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0=35\mathbf{i}-30\mathbf{j}

The next stage has

\displaystyle t=40, \displaystyle \mathbf{a}=0, \displaystyle {{\mathbf{r}}_{0}}=35\mathbf{i}-30\mathbf{j}, and \displaystyle \mathbf{u}=6\mathbf{i}-8\mathbf{j} as the final position and velocity of the first stage is the initial position and velocity of the second stage.

Using once again \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} we get

\displaystyle \mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+0+(35\mathbf{i}-30\mathbf{j})=275\mathbf{i}-350\mathbf{j}

This is the boat´s final position.

The distance from the starting point is the magnitude of this vector,


\displaystyle \sqrt{{{275}^{2}}+{{\left( -355 \right)}^{2}}}=445\ \text{m}

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