Solution 8.6b
From Mechanics
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We aassume the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math> | We aassume the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math> | ||
| - | The boat first accelerates to a point <math>A</math> say. We first must calculate the position of this point <math></math>. | + | 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>. |
Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with, | Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with, | ||
<math>t=10</math>, <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> and from part a) | <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> | <math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math> | ||
| - | + | <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</math | |
Revision as of 18:19, 15 April 2010
We aassume 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
