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Revision as of 15:18, 18 April 2010 (edit) Ian (Talk | contribs) (New page: We must find the position of the particle after 30 s. Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with <math>\ \mathbf{a}=\mathbf{i}+2\...) ← Previous diff		Current revision (15:39, 18 April 2010) (edit) (undo) Ian (Talk | contribs)
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	Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>		Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>
-	with <math>\ \mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}}</math>,	+	with <math>\ \mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}}</math> which was calculated in part a),
-	<math>\ \mathbf{u}=<math>4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> and	+	<math>\ \mathbf{u}=4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> and
-	<math>{{\mathbf{r}}_{0}}=0</math> gives,	+	<math>{{\ \mathbf{r}}_{0}}=0</math> gives,
-	<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>	+	<math>\mathbf{r}=(4\mathbf{i}+6\mathbf{j}) \times 30+\frac{1}{2}(\mathbf{i}+2\mathbf{j}) \times {{30}^{\ 2}}</math>
		+
		+	<math>\mathbf{r}=120\mathbf{i}+180\mathbf{j} +450\mathbf{i}+900\mathbf{j} = 570\mathbf{i}+1080\mathbf{j}</math>
		+
		+	As the starting point is at the origin, the distance is the magnitude of this position vector.
		+
		+	<math>\sqrt{{{570}^{2}}+{{108}^{2}}}=\sqrt{324900+1166400}=\sqrt{1491300}=1221\ \text{m}</math>

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Current revision

We must find the position of the particle after 30 s.

Using \displaystyle \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} with \displaystyle \ \mathbf{a}=\mathbf{i}+2\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}} which was calculated in part a), \displaystyle \ \mathbf{u}=4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}} and \displaystyle {{\ \mathbf{r}}_{0}}=0 gives,

\displaystyle \mathbf{r}=(4\mathbf{i}+6\mathbf{j}) \times 30+\frac{1}{2}(\mathbf{i}+2\mathbf{j}) \times {{30}^{\ 2}}

\displaystyle \mathbf{r}=120\mathbf{i}+180\mathbf{j} +450\mathbf{i}+900\mathbf{j} = 570\mathbf{i}+1080\mathbf{j}

As the starting point is at the origin, the distance is the magnitude of this position vector.

\displaystyle \sqrt{{{570}^{2}}+{{108}^{2}}}=\sqrt{324900+1166400}=\sqrt{1491300}=1221\ \text{m}