From Mechanics

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-	Here we use	+	We aassume the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math>
-	<math>\mathbf{r}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t+{{\mathbf{r}}_{0}}</math>	+	The boat first accelerates to a point <math>A</math> say. We first must calculate the position of this point <math></math>.
-	in the first part.	+	Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with,
-	According to the text	+	<math>t=10</math>, <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> and from part a)
-	<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>	+	<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>
-	<math>\mathbf{v}=6\mathbf{i}-8\mathbf{j}</math>	+	Using <math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math
-		+
-	We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.	+
-		+
-	At <math>t=10</math>	+
-		+
-	<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>	+
-		+
-	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 \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 .

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}}

Using \displaystyle {{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}