Solution 8.8b - Revision history

Ian at 15:39, 18 April 2010

Ian — Sun, 18 Apr 2010 15:39:04 GMT

←Older revision		Revision as of 15:39, 18 April 2010
Line 8:		Line 8:
	<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}=(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} = ~~470~~\mathbf{i}+1080\mathbf{j}</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>

Ian at 15:26, 18 April 2010

Ian — Sun, 18 Apr 2010 15:26:27 GMT

←Older revision		Revision as of 15:26, 18 April 2010
Line 7:		Line 7:

	<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}=(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} = 470\mathbf{i}+1080\mathbf{j}</math>

Ian at 15:23, 18 April 2010

Ian — Sun, 18 Apr 2010 15:23:26 GMT

←Older revision		Revision as of 15:23, 18 April 2010
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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>

Ian: New page: We must find the position of the particle after 30 s. Using $\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}$ with $\ \mathbf{a}=\mathbf{i}+2\...$

Ian — Sun, 18 Apr 2010 15:18:40 GMT

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\...

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\mathbf{j} \text{ m}{{\text{s}}^{\text{-2}}}</math>,
<math>\ \mathbf{u}=<math>4\mathbf{i}+6\mathbf{j}\text{ m}{{\text{s}}^{\text{-1}}}</math> and
<math>{{\mathbf{r}}_{0}}=0</math> gives,

<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>

Solution 8.8b - Revision history

Ian at 15:39, 18 April 2010

Ian at 15:26, 18 April 2010

Ian at 15:23, 18 April 2010

Ian: New page: We must find the position of the particle after 30 s. Using \mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} with \ \mathbf{a}=\mathbf{i}+2\...

Ian: New page: We must find the position of the particle after 30 s. Using $\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}$ with $\ \mathbf{a}=\mathbf{i}+2\...$