Solution 8.6b
From Mechanics
| Line 8: | Line 8: | ||
<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 | + | <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}+70\mathbf{j}</math> |
| + | |||
| + | The next stage has | ||
| + | |||
| + | <math>t=40</math>, <math>\mathbf{a}=0</math>, <math>{{\mathbf{r}}_{0}}=35\mathbf{i}+70\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. | ||
| + | |||
| + | Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get | ||
Revision as of 18:30, 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=35\mathbf{i}+70\mathbf{j}
The next stage has
\displaystyle t=40, \displaystyle \mathbf{a}=0, \displaystyle {{\mathbf{r}}_{0}}=35\mathbf{i}+70\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
