Solution 8.6b
From Mechanics
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Here we use | Here we use | ||
- | <math>\mathbf{r}= | + | <math>\mathbf{r}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t+{{\mathbf{r}}_{0}}</math> |
+ | |||
+ | in the first part. | ||
According to the text | According to the text | ||
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<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> | <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> | ||
- | + | <math>\mathbf{v}=6\mathbf{i}-8\mathbf{j}</math> | |
- | + | ||
- | <math>\mathbf{ | + | |
We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>. | We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>. | ||
- | At <math>t=10 | + | At <math>t=10</math> |
- | <math>\mathbf{r}=(\mathbf{i}+2\mathbf{j} | + | <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> |
+ | The distance is the magnitude of this vector | ||
+ | <math>\sqrt{{{7}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{13}=3\textrm{.}6\ \text{m}</math> | ||
- | + | Thus during the first part the boat has travelled a distance 3.6 m. |
Revision as of 19:54, 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}
Thus during the first part the boat has travelled a distance 3.6 m.