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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		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}=(6\mathbf{i}-8\mathbf{j}) \times 40+35\mathbf{i}+70\mathbf{j}=275\mathbf{i}-250\mathbf{j}</math>

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Revision as of 18:36, 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

\displaystyle \mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+35\mathbf{i}+70\mathbf{j}=275\mathbf{i}-250\mathbf{j}