Solution 8.4a - Revision history

Ian at 12:17, 27 March 2011

2011-03-27T12:17:43Z

←Older revision		Revision as of 12:17, 27 March 2011
Line 9:		Line 9:
	The speed is the magnitude of this velocity vector.		The speed is the magnitude of this velocity vector.

-	<math>v=\sqrt{{{6}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{61}=7\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>	+	<math>v=\sqrt{{{6}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{61}=7\textrm{.}81\ \text{m}{{\text{s}}^{-1}}</math>

Ian at 16:05, 13 April 2010

2010-04-13T16:05:09Z

←Older revision		Revision as of 16:05, 13 April 2010
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-	Here we have <math>\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\ \ \mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}</math>.	+	Here we have <math>\mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\ \ \mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}</math>.

	Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t\ \</math> at <math>\ \ t=10</math> gives		Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t\ \</math> at <math>\ \ t=10</math> gives

-	<math>\mathbf{v}=4\mathbf{i}– 8\mathbf{j}+(0.2\mathbf{i}+0.3\mathbf{j})\times 10\ \</math> or	+	<math>\mathbf{v}=4\mathbf{i}– 8\mathbf{j}+(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j})\times 10\ \</math> or

-	<math>\mathbf{v}=6\mathbf{i}– 5\mathbf{j}</math>	+	<math>\mathbf{v}=6\mathbf{i}– 5\mathbf{j}\ \ \text{m}{{\text{s}}^{-1}}</math>

		+	The speed is the magnitude of this velocity vector.

-		+	<math>v=\sqrt{{{6}^{2}}+{{\left( -5 \right)}^{2}}}=\sqrt{61}=7\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>
-		+
-		+
-	~~Using~~ <math>\~~mathbf~~{r}~~=\mathbf~~{u}t+~~\frac~~{1}{2}\~~mathbf~~{a}{{t}^{~~\ 2}~~}+{{\~~mathbf~~{r}}_{0}}</math>	+

Ian at 15:57, 13 April 2010

2010-04-13T15:57:18Z

←Older revision		Revision as of 15:57, 13 April 2010
Line 2:		Line 2:

	Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t\ \</math> at <math>\ \ t=10</math> gives		Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t\ \</math> at <math>\ \ t=10</math> gives
		+
		+	<math>\mathbf{v}=4\mathbf{i}– 8\mathbf{j}+(0.2\mathbf{i}+0.3\mathbf{j})\times 10\ \</math> or
		+
		+	<math>\mathbf{v}=6\mathbf{i}– 5\mathbf{j}</math>

Ian at 15:48, 13 April 2010

2010-04-13T15:48:42Z

←Older revision		Revision as of 15:48, 13 April 2010
Line 1:		Line 1:
-	Here we have <math>\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}</math>.	+	Here we have <math>\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\ \ \mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}</math>.
		+
		+	Using <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t\ \</math> at <math>\ \ t=10</math> gives
		+
		+
		+
		+
		+
		+	Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>

Ian at 15:43, 13 April 2010

2010-04-13T15:43:56Z

←Older revision		Revision as of 15:43, 13 April 2010
Line 1:		Line 1:
-	Here we have <math>\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>{\mathbf{ r}}_{0}=6\mathbf{i}+2\mathbf{j}</math>.	+	Here we have <math>\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>\mathbf{ r}_{0}=6\mathbf{i}+2\mathbf{j}</math>.

Ian: New page: Here we have $\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \$ $\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \$ and ${\mathbf{ r}}_{0}=6\mathbf{i}+2\mathbf{j}$ .

2010-04-13T15:42:50Z

New page: Here we have <math>\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>{\mathbf{ r}}_{0}=6\mathbf{i}+2\mathbf{j}</math>.

New page

Here we have <math>\mathbf{a}=0.2\mathbf{i}+0.3\mathbf{j},\ \</math> <math>\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \</math> and <math>{\mathbf{ r}}_{0}=6\mathbf{i}+2\mathbf{j}</math>.