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Solution 8.6b

From Mechanics

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Here we use
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We aassume 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></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{u}=\mathbf{i}+2\mathbf{j}</math>
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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{v}=6\mathbf{i}-8\mathbf{j}</math>
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Using <math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math
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We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.
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At <math>t=10</math>
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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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The distance is the magnitude of this vector
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<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.
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Revision as of 18:15, 15 April 2010

We aassume the starting point is the origin. This means r0=0

The boat first accelerates to a point A say. We first must calculate the position of this point .

Using r=ut+21at 2+r0 with,

t=10, u=i+2j and from part a)

a=0.5i−j ms−2

Using rA=(i+2j)10+21at 2+r0

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