Solution 8.10a - Revision history

Ian at 18:50, 19 April 2010

2010-04-19T18:50:11Z

←Older revision		Revision as of 18:50, 19 April 2010
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	We can now substitute in the equation <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> giving,		We can now substitute in the equation <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> giving,

-	<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>	+	<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>.
		+
		+	We are looking for the time when the height is <math>90\ \text{m}</math>, that is when the <math>\mathbf{k}</math> component is <math>80\ \text{m}</math>. This gives,
		+
		+	<math>\frac{1}{2} \times 0\textrm{.}1{{t}^{\ 2}}+10=90</math>
		+
		+	or
		+
		+	<math>\begin{align}
		+	& \frac{1}{20}{{t}^{\ 2}}=80 \\
		+	& \\
		+	& {{t}^{\ 2}}=1600 \\
		+	& \\
		+	& t=40\ \text{s} \\
		+	\end{align}</math>

Ian at 18:25, 19 April 2010

2010-04-19T18:25:16Z

←Older revision		Revision as of 18:25, 19 April 2010
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	<math>{{\mathbf{r}}_{0}}=10\mathbf{k}\text{ m}</math>		<math>{{\mathbf{r}}_{0}}=10\mathbf{k}\text{ m}</math>

-	<math>\mathbf{a}=c\text{ m}{{\text{s}}^{\text{-2}}}</math>	+	<math>\mathbf{a}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}^{\text{-2}}}</math>

	We need to calculate <math>\mathbf{u}</math>.		We need to calculate <math>\mathbf{u}</math>.

Ian at 17:42, 19 April 2010

2010-04-19T17:42:24Z

←Older revision		Revision as of 17:42, 19 April 2010
Line 3:		Line 3:
	<math>{{\mathbf{r}}_{0}}=10\mathbf{k}\text{ m}</math>		<math>{{\mathbf{r}}_{0}}=10\mathbf{k}\text{ m}</math>

-	<math>\mathbf{a}=~~(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})~~\text{ m}{{\text{s}}^{\text{-2}}}</math>	+	<math>\mathbf{a}=c\text{ m}{{\text{s}}^{\text{-2}}}</math>

	We need to calculate <math>\mathbf{u}</math>.		We need to calculate <math>\mathbf{u}</math>.
Line 21:		Line 21:

	We can now substitute in the equation <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> giving,		We can now substitute in the equation <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> giving,
		+
		+	<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>

Ian at 16:32, 19 April 2010

2010-04-19T16:32:43Z

←Older revision		Revision as of 16:32, 19 April 2010
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	<math>\mathbf{a}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}^{\text{-2}}}</math>		<math>\mathbf{a}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}^{\text{-2}}}</math>
		+
		+	We need to calculate <math>\mathbf{u}</math>.

	If a vector, like the starting velocity <math>\mathbf{u}</math> in this problem, points north east this means it has the same component in the east and north directions. In other words its <math>\mathbf{i}</math> component and its <math>\mathbf{j}</math> component are the same. Thus <math>\mathbf{u}</math> is of the form		If a vector, like the starting velocity <math>\mathbf{u}</math> in this problem, points north east this means it has the same component in the east and north directions. In other words its <math>\mathbf{i}</math> component and its <math>\mathbf{j}</math> component are the same. Thus <math>\mathbf{u}</math> is of the form
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	<math>\mathbf{u}=8\ \mathbf{i}+8 \ \mathbf{ j}\text{ m}{{\text{s}}^{\text{-1}}}</math>		<math>\mathbf{u}=8\ \mathbf{i}+8 \ \mathbf{ j}\text{ m}{{\text{s}}^{\text{-1}}}</math>
		+
		+	We can now substitute in the equation <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> giving,

Ian at 16:30, 19 April 2010

2010-04-19T16:30:07Z

←Older revision		Revision as of 16:30, 19 April 2010
Line 10:		Line 10:

	This has magnitude		This has magnitude
-	<math>\sqrt{{{B}^{2}}+{{B}^{2}}}=\sqrt{2}B</math>	+	<math>\sqrt{{{B}^{2}}+{{B}^{2}}}=\sqrt{2}B \text{ m}{{\text{s}}^{\text{-1}}}</math>

	However the magnitude of the initial velocity is the initial speed which is		However the magnitude of the initial velocity is the initial speed which is

Ian at 16:28, 19 April 2010

2010-04-19T16:28:50Z

←Older revision		Revision as of 16:28, 19 April 2010
Line 12:		Line 12:
	<math>\sqrt{{{B}^{2}}+{{B}^{2}}}=\sqrt{2}B</math>		<math>\sqrt{{{B}^{2}}+{{B}^{2}}}=\sqrt{2}B</math>

-	However the magnitude of the initial velocity is the initial ~~sped~~ which is	+	However the magnitude of the initial velocity is the initial speed which is

-	<math>8\sqrt{2}</math>.	+	<math>8\sqrt{2}\text{ m}{{\text{s}}^{\text{-1}}}</math>. Comparing these results shows <math>B=8</math> and thus
		+
		+	<math>\mathbf{u}=8\ \mathbf{i}+8 \ \mathbf{ j}\text{ m}{{\text{s}}^{\text{-1}}}</math>

Ian: New page: We use, $\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}$ to find an expression for the position of the aeroplane at a time t. Here, ${{\mathbf{...$

2010-04-19T16:25:43Z

New page: We use, <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> to find an expression for the position of the aeroplane at a time t. Here, <math>{{\mathbf{...

New page

We use, <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> to find an expression for the position of the aeroplane at a time t. Here,

<math>{{\mathbf{r}}_{0}}=10\mathbf{k}\text{ m}</math>

<math>\mathbf{a}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}^{\text{-2}}}</math>

If a vector, like the starting velocity <math>\mathbf{u}</math> in this problem, points north east this means it has the same component in the east and north directions. In other words its <math>\mathbf{i}</math> component and its <math>\mathbf{j}</math> component are the same. Thus <math>\mathbf{u}</math> is of the form

<math>\mathbf{u}=B\ \mathbf{i}+B \ \mathbf{ j}</math>

This has magnitude
<math>\sqrt{{{B}^{2}}+{{B}^{2}}}=\sqrt{2}B</math>

However the magnitude of the initial velocity is the initial sped which is

<math>8\sqrt{2}</math>.