Ian at 14:18, 14 April 2010

Ian — Wed, 14 Apr 2010 14:18:24 GMT

←Older revision		Revision as of 14:18, 14 April 2010
Line 5:		Line 5:
	Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</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> at <math>\ \ t=10</math> gives

-	~~Using~~ <math>\mathbf{r}=(4\mathbf{i}– 8\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j}) \times {{10}^{\ 2}}+6\mathbf{i}+2\mathbf{j}</math>	+	<math>\mathbf{r}=(4\mathbf{i}– 8\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j}) \times {{10}^{\ 2}}+6\mathbf{i}+2\mathbf{j}</math>

	<math>\mathbf{r}=56\mathbf{i}– 63\mathbf{j}\ \ \text{m}{{\text{s}}^{-1}}</math>		<math>\mathbf{r}=56\mathbf{i}– 63\mathbf{j}\ \ \text{m}{{\text{s}}^{-1}}</math>

Ian: New page: Here we have $\mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \$ $\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \$ and $\ \ \mathbf{ r}_{0}=6\mathbf{i}...$

Ian — Tue, 13 Apr 2010 16:19:33 GMT

New page: 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}...

New page

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{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> at <math>\ \ t=10</math> gives

Using <math>\mathbf{r}=(4\mathbf{i}– 8\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j}) \times {{10}^{\ 2}}+6\mathbf{i}+2\mathbf{j}</math>

<math>\mathbf{r}=56\mathbf{i}– 63\mathbf{j}\ \ \text{m}{{\text{s}}^{-1}}</math>

Solution 8.4b - Revision history

Ian at 14:18, 14 April 2010

Ian: New page: Here we have \mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \ \mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \ and \ \ \mathbf{ r}_{0}=6\mathbf{i}...

Ian: New page: Here we have $\mathbf{a}=0\textrm{.}2\mathbf{i}+0\textrm{.}3\mathbf{j},\ \$ $\mathbf{u}=4\mathbf{i}– 8\mathbf{j}\ \$ and $\ \ \mathbf{ r}_{0}=6\mathbf{i}...$