Solution 8.6b - Revision history

Ian at 12:31, 27 March 2011

Ian — Sun, 27 Mar 2011 12:31:00 GMT

←Older revision		Revision as of 12:31, 27 March 2011
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	<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>		<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>

-	<math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0=35\mathbf{i}-50\mathbf{j}</math>	+	<math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0=35\mathbf{i}-30\mathbf{j}</math>

	The next stage has		The next stage has

-	<math>t=40</math>, <math>\mathbf{a}=0</math>, <math>{{\mathbf{r}}_{0}}=35\mathbf{i}~~+70~~\mathbf{j}</math>, and <math>\mathbf{u}=6\mathbf{i}-8\mathbf{j}</math> as the final position and velocity of the first stage is the initial position and velocity of the second stage.	+	<math>t=40</math>, <math>\mathbf{a}=0</math>, <math>{{\mathbf{r}}_{0}}=35\mathbf{i}-30\mathbf{j}</math>, and <math>\mathbf{u}=6\mathbf{i}-8\mathbf{j}</math> as the final position and velocity of the first stage is the initial position and velocity of the second stage.

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

-	<math>\mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+0+(35\mathbf{i}-50\mathbf{j})=275\mathbf{i}-~~355~~\mathbf{j}</math>	+	<math>\mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+0+(35\mathbf{i}-30\mathbf{j})=275\mathbf{i}-350\mathbf{j}</math>

	This is the boat´s final position.		This is the boat´s final position.
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-	<math>\sqrt{{{275}^{2}}+{{\left( -355 \right)}^{2}}}=~~450~~\ \text{m}</math>	+	<math>\sqrt{{{275}^{2}}+{{\left( -355 \right)}^{2}}}=445\ \text{m}</math>

Ian at 19:42, 15 April 2010

Ian — Thu, 15 Apr 2010 19:42:33 GMT

←Older revision		Revision as of 19:42, 15 April 2010
Line 1:		Line 1:
-	We ~~aassume~~ the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math>	+	We assume the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math>

	The boat first accelerates to a point <math>A</math> say. We first must calculate the position of this point <math>{{\mathbf{r}}_{A}}</math>.		The boat first accelerates to a point <math>A</math> say. We first must calculate the position of this point <math>{{\mathbf{r}}_{A}}</math>.
Line 8:		Line 8:
	<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>		<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>

-	<math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0=35\mathbf{i}~~+70~~\mathbf{j}</math>	+	<math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0=35\mathbf{i}-50\mathbf{j}</math>

	The next stage has		The next stage has
Line 16:		Line 16:
	Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get		Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get

-	<math>\mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+35\mathbf{i}~~+70~~\mathbf{j}=275\mathbf{i}-~~250~~\mathbf{j}</math>	+	<math>\mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+0+(35\mathbf{i}-50\mathbf{j})=275\mathbf{i}-355\mathbf{j}</math>
		+
		+	This is the boat´s final position.
		+
		+	The distance from the starting point is the magnitude of this vector,
		+
		+
		+	<math>\sqrt{{{275}^{2}}+{{\left( -355 \right)}^{2}}}=450\ \text{m}</math>

Ian at 18:36, 15 April 2010

Ian — Thu, 15 Apr 2010 18:36:20 GMT

←Older revision		Revision as of 18:36, 15 April 2010
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	Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get		Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get
		+
		+	<math>\mathbf{r}=(6\mathbf{i}-8\mathbf{j}) \times 40+35\mathbf{i}+70\mathbf{j}=275\mathbf{i}-250\mathbf{j}</math>

Ian at 18:30, 15 April 2010

Ian — Thu, 15 Apr 2010 18:30:51 GMT

←Older revision		Revision as of 18:30, 15 April 2010
Line 8:		Line 8:
	<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>		<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>

-	<math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0</math	+	<math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0=35\mathbf{i}+70\mathbf{j}</math>
		+
		+	The next stage has
		+
		+	<math>t=40</math>, <math>\mathbf{a}=0</math>, <math>{{\mathbf{r}}_{0}}=35\mathbf{i}+70\mathbf{j}</math>, and <math>\mathbf{u}=6\mathbf{i}-8\mathbf{j}</math> as the final position and velocity of the first stage is the initial position and velocity of the second stage.
		+
		+	Using once again <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> we get

Ian at 18:19, 15 April 2010

Ian — Thu, 15 Apr 2010 18:19:27 GMT

←Older revision		Revision as of 18:19, 15 April 2010
Line 1:		Line 1:
	We aassume the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math>		We aassume the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math>

-	The boat first accelerates to a point <math>A</math> say. We first must calculate the position of this point <math></math>.	+	The boat first accelerates to a point <math>A</math> say. We first must calculate the position of this point <math>{{\mathbf{r}}_{A}}</math>.

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

	<math>t=10</math>, <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> and from part a)		<math>t=10</math>, <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> and from part a)
-
	<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>		<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>

-	~~Using~~ <math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+~~{{\mathbf{r}}_{~~0}}</math	+	<math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}(0\textrm{.}5\mathbf{i}-\mathbf{j}) \times {{10}^{\ 2}}+0</math

Ian at 18:15, 15 April 2010

Ian — Thu, 15 Apr 2010 18:15:00 GMT

←Older revision		Revision as of 18:15, 15 April 2010
Line 1:		Line 1:
-	~~Here we use~~	+	We aassume the starting point is the origin. This means <math>{{\mathbf{r}}_{0}}=0</math>

-	<math>~~\mathbf{r}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t+{{\mathbf{r}}_{0}}~~</math>	+	The boat first accelerates to a point <math>A</math> say. We first must calculate the position of this point <math></math>.

-	~~in the first part.~~	+	Using <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math> with,

-	~~According to the text~~	+	<math>t=10</math>, <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> and from part a)

-	<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>	+	<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}\ \text{m}{{\text{s}}^{-2}}</math>

-	~~<math>\mathbf{v}=6\mathbf{i}-8\mathbf{j}</math>~~	+	Using <math>{{\mathbf{r}}_{A}}=(\mathbf{i}+2\mathbf{j}) \times 10+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math
-		+
-	~~We assume the starting point is the origin so that~~ <math>{{\mathbf{r}}_{0}}=~~0</math>.~~	+
-		+
-	~~At <math>t=10</math>~~	+
-		+
-	~~<math> \mathbf{r}=\frac{1}{2}~~(\mathbf{i}+2~~\mathbf{j}+6\mathbf{i}-8\mathbf{j})t=(3\textrm{.}5\mathbf{i}-3~~\mathbf{j}) \times 10~~=35~~\~~mathbf~~{i}~~-30\mathbf~~{j}~~=(7~~\mathbf{i}~~-6\mathbf~~{~~j}) \times 5</math>~~	+
-		+
-	~~The distance is the magnitude of this vector~~	+
-		+
-	~~<math>\sqrt~~{~~{{7~~}^{2}}+{{\~~left( -6 \right)}^~~{2}}~~}=\sqrt~~{13}~~=3\textrm{.}6\ \text{m~~}</math>	+
-		+
-	~~In the second part the boat travel with constant velocity <math>6\mathbf{i}-8\mathbf{j}</math>~~	+
-		+
-		+
-		+
-		+
-		+
-	~~Thus during the first part the boat has travelled a distance 3.6 m.~~	+

Ian at 19:57, 14 April 2010

Ian — Wed, 14 Apr 2010 19:57:27 GMT

←Older revision		Revision as of 19:57, 14 April 2010
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	<math>\sqrt{{{7}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{13}=3\textrm{.}6\ \text{m}</math>		<math>\sqrt{{{7}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{13}=3\textrm{.}6\ \text{m}</math>
		+
		+	In the second part the boat travel with constant velocity <math>6\mathbf{i}-8\mathbf{j}</math>
		+
		+
		+
		+

	Thus during the first part the boat has travelled a distance 3.6 m.		Thus during the first part the boat has travelled a distance 3.6 m.

Ian at 19:54, 14 April 2010

Ian — Wed, 14 Apr 2010 19:54:07 GMT

←Older revision		Revision as of 19:54, 14 April 2010
Line 1:		Line 1:
	Here we use		Here we use

-	<math>\mathbf{r}=~~\mathbf{u}t+~~\frac{1}{2}\mathbf{a}{{t~~}^{\ 2}}~~+{{\mathbf{r}}_{0}} </math>	+	<math>\mathbf{r}=\frac{1}{2}(\mathbf{u}+\mathbf{v})t+{{\mathbf{r}}_{0}}</math>
		+
		+	in the first part.

	According to the text		According to the text
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	<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>		<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>

-	~~and part a) gave~~	+	<math>\mathbf{v}=6\mathbf{i}-8\mathbf{j}</math>
-		+
-	<math>\mathbf{a}=~~0\textrm{.}5~~\mathbf{i}-\mathbf{j}</math>	+

	We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.		We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.

-	At <math>t=10~~+40=50~~</math> ~~we obtain~~	+	At <math>t=10</math>

-	<math>\mathbf{r}=(\mathbf{i}+2\mathbf{j}~~) \times 50~~+\~~frac~~{1}~~{2}(~~\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j} )\times{{50}^{\ 2}}</math>	+	<math> \mathbf{r}=\frac{1}{2}(\mathbf{i}+2\mathbf{j}+6\mathbf{i}-8\mathbf{j})t=(3\textrm{.}5\mathbf{i}-3\mathbf{j}) \times 10=35\mathbf{i}-30\mathbf{j}=(7\mathbf{i}-6\mathbf{j}) \times 5</math>

		+	The distance is the magnitude of this vector

		+	<math>\sqrt{{{7}^{2}}+{{\left( -6 \right)}^{2}}}=\sqrt{13}=3\textrm{.}6\ \text{m}</math>

-	~~The distance is~~ the ~~magnitude of this vector~~	+	Thus during the first part the boat has travelled a distance 3.6 m.

Ian at 19:35, 14 April 2010

Ian — Wed, 14 Apr 2010 19:35:10 GMT

←Older revision		Revision as of 19:35, 14 April 2010
Line 7:		Line 7:
	<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>		<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>

		+	and part a) gave
		+
		+	<math>\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j}</math>

	We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.		We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.
Line 12:		Line 15:
	At <math>t=10+40=50</math> we obtain		At <math>t=10+40=50</math> we obtain

-		+	<math>\mathbf{r}=(\mathbf{i}+2\mathbf{j}) \times 50+\frac{1}{2}(\mathbf{a}=0\textrm{.}5\mathbf{i}-\mathbf{j} )\times{{50}^{\ 2}}</math>



	The distance is the magnitude of this vector		The distance is the magnitude of this vector

Ian: New page: Here we use $\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}$ According to the text $\mathbf{u}=\mathbf{i}+2\mathbf{j}$ We assume th...

Ian — Wed, 14 Apr 2010 19:31:44 GMT

New page: Here we use <math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}} </math> According to the text <math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math> We assume th...

New page

Here we use

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

According to the text

<math>\mathbf{u}=\mathbf{i}+2\mathbf{j}</math>

We assume the starting point is the origin so that <math>{{\mathbf{r}}_{0}}=0</math>.

At <math>t=10+40=50</math> we obtain

The distance is the magnitude of this vector