Solution 8.1a - Revision history

Ian at 12:16, 27 March 2011

Ian — Sun, 27 Mar 2011 12:16:10 GMT

←Older revision		Revision as of 12:16, 27 March 2011
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	which gives		which gives

-	<math>\mathbf{v}=13\mathbf{i}+9\mathbf{j} \ \text{ m}{{\text{s}}^{\text{-1}}}</math>	+	<math>\mathbf{v}=13\mathbf{i}+7\mathbf{j} \ \text{ m}{{\text{s}}^{\text{-1}}}</math>

Ian at 15:27, 13 April 2010

Ian — Tue, 13 Apr 2010 15:27:56 GMT

←Older revision		Revision as of 15:27, 13 April 2010
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	Substituting these into the equation		Substituting these into the equation
	<math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ \</math>		<math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ \</math>
-	gives the velocity of the ~~particle~~ at time <math>t=10 \text{ s }</math> as:	+	gives the velocity of the jet-ski at time <math>t=10 \text{ s }</math> as:

	<math>\mathbf{v}=4\mathbf{i}+(0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j})\times 10</math>		<math>\mathbf{v}=4\mathbf{i}+(0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j})\times 10</math>

Ian at 15:26, 13 April 2010

Ian — Tue, 13 Apr 2010 15:26:53 GMT

←Older revision		Revision as of 15:26, 13 April 2010
Line 6:		Line 6:
	Substituting these into the equation		Substituting these into the equation
	<math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ \</math>		<math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ \</math>
-	gives the velocity of the ~~ball~~ at time <math>t=10 \text{ s }</math> as:	+	gives the velocity of the particle at time <math>t=10 \text{ s }</math> as:

	<math>\mathbf{v}=4\mathbf{i}+(0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j})\times 10</math>		<math>\mathbf{v}=4\mathbf{i}+(0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j})\times 10</math>

Ian at 17:44, 12 April 2010

Ian — Mon, 12 Apr 2010 17:44:07 GMT

←Older revision	Revision as of 17:44, 12 April 2010
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	+	In this case we have
	+	<math>\mathbf{u}=4\mathbf{i}</math>
	+	and
	+	<math>\mathbf{a}=0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j}</math>

	+	Substituting these into the equation
	+	<math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ \</math>
	+	gives the velocity of the ball at time <math>t=10 \text{ s }</math> as:
	+
	+	<math>\mathbf{v}=4\mathbf{i}+(0\textrm{.}9\mathbf{i}+0\textrm{.}7\mathbf{j})\times 10</math>
	+
	+	which gives
	+
	+	<math>\mathbf{v}=13\mathbf{i}+9\mathbf{j} \ \text{ m}{{\text{s}}^{\text{-1}}}</math>

Ian: Removing all content from page

Ian — Mon, 12 Apr 2010 17:24:22 GMT

Removing all content from page

←Older revision		Revision as of 17:24, 12 April 2010
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-	In this case we have
-	<math>\mathbf{u}=4\mathbf{i}</math>
-	,
-	<math>\mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j}</math>
-	and
-	<math>{{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}</math>.

-	Substituting these into the equation
-	<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+400\mathbf{i}+350\mathbf{j}</math>
-	gives the position vector of the ball at time <math>10</math> as:
-
-	<math>\begin{align}
-	& \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\
-	& =40\mathbf{i}+45\mathbf{i}+35\mathbf{j}+400\mathbf{i}+350\mathbf{j} \\ \\
-	& =485\mathbf{i}+385\mathbf{j} \ \text{m}
-	\end{align}</math>

Ian at 17:20, 12 April 2010

Ian — Mon, 12 Apr 2010 17:20:53 GMT

←Older revision		Revision as of 17:20, 12 April 2010
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	<math>\begin{align}		<math>\begin{align}
	& \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\		& \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\
-	& =8t\mathbf{i}+2t\mathbf{~~j}-5{{t}^{\ 2}~~}\mathbf{j}+1\~~textrm~~{.}5\mathbf{j} \\	+	& =40\mathbf{i}+45\mathbf{i}+35\mathbf{j}+400\mathbf{i}+350\mathbf{j} \\ \\
-	& =~~(8t)~~\mathbf{i}+~~(1\textrm{.}5+2t-5{{t}^{\ 2}})~~\mathbf{j} \ \text{m}	+	& =485\mathbf{i}+385\mathbf{j} \ \text{m}
	\end{align}</math>		\end{align}</math>

Ian at 17:15, 12 April 2010

Ian — Mon, 12 Apr 2010 17:15:12 GMT

←Older revision		Revision as of 17:15, 12 April 2010
Line 7:		Line 7:

	Substituting these into the equation		Substituting these into the equation
-	<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>	+	<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+400\mathbf{i}+350\mathbf{j}</math>
	gives the position vector of the ball at time <math>10</math> as:		gives the position vector of the ball at time <math>10</math> as:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{r}=(4\mathbf{i})t+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j}){{t}^{\ 2}}+1\~~textrm~~{.}5\mathbf{j} \\	+	& \mathbf{r}=(4\mathbf{i})\times 10+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j})\times {{10}^{\ 2}}+400\mathbf{i}+350\mathbf{j} \\
	& =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\		& =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\
	& =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m}		& =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m}
	\end{align}</math>		\end{align}</math>

Ian: New page: In this case we have $\mathbf{u}=4\mathbf{i}$ , $\mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j}$ and ${{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}$ . Substit...

Ian — Mon, 12 Apr 2010 17:10:51 GMT

New page: In this case we have <math>\mathbf{u}=4\mathbf{i}</math> , <math>\mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j}</math> and <math>{{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}</math>. Substit...

New page

In this case we have
<math>\mathbf{u}=4\mathbf{i}</math>
,
<math>\mathbf{a}=0.9\mathbf{i}+0.7\mathbf{j}</math>
and
<math>{{\mathbf{r}}_{0}}=400\mathbf{i}+350\mathbf{j}</math>.

Substituting these into the equation
<math>\mathbf{r}=\mathbf{u}t+\frac{1}{2}\mathbf{a}{{t}^{\ 2}}+{{\mathbf{r}}_{0}}</math>
gives the position vector of the ball at time <math>10</math> as:

<math>\begin{align}
& \mathbf{r}=(4\mathbf{i})t+\frac{1}{2}(0.9\mathbf{i}+0.7\mathbf{j}){{t}^{\ 2}}+1\textrm{.}5\mathbf{j} \\
& =8t\mathbf{i}+2t\mathbf{j}-5{{t}^{\ 2}}\mathbf{j}+1\textrm{.}5\mathbf{j} \\
& =(8t)\mathbf{i}+(1\textrm{.}5+2t-5{{t}^{\ 2}})\mathbf{j} \ \text{m}
\end{align}</math>