Ian at 12:31, 20 April 2010

Ian — Tue, 20 Apr 2010 12:31:22 GMT

←Older revision		Revision as of 12:31, 20 April 2010
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	<math>\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j}) \times 40+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}) \times{{40}^{\ 2}}+10\mathbf{k}=328\mathbf{i}+336\mathbf{j}+10\mathbf{k} \ </math>, for any time <math>t</math>.		<math>\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j}) \times 40+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}) \times{{40}^{\ 2}}+10\mathbf{k}=328\mathbf{i}+336\mathbf{j}+10\mathbf{k} \ </math>, for any time <math>t</math>.
		+
		+	The car starts at the origin and finishes at <math>328\mathbf{i}+336\mathbf{j}</math>.
		+
		+	The magnitude of this position vector is the distance <math>D\ </math> travelled.
		+
		+
		+	<math>D=\sqrt{{{328}^{2}}+{{336}^{2}}}=\sqrt{107584+112896}=\sqrt{220480}=470\ \text{m}</math>

Ian: New page: We use the result from part a). $\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{...$

Ian — Tue, 20 Apr 2010 12:25:34 GMT

New page: We use the result from part a). <math>\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{...

New page

We use the result from part a).

<math>\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{k}\ </math>, for any time <math>t</math>.

At the end when <math>t=40</math> also from part a), we get

<math>\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j}) \times 40+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}) \times{{40}^{\ 2}}+10\mathbf{k}=328\mathbf{i}+336\mathbf{j}+10\mathbf{k} \ </math>, for any time <math>t</math>.

Solution 8.10c - Revision history

Ian at 12:31, 20 April 2010

Ian: New page: We use the result from part a). \mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{...

Ian: New page: We use the result from part a). $\mathbf{r}=(8\ \mathbf{i}+8 \ \mathbf{ j})t+\frac{1}{2}(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}){{t}^{\ 2}}+10\mathbf{...$