Solution 8.10b - Revision history

Ian at 12:45, 14 May 2010

2010-05-14T12:45:24Z

←Older revision		Revision as of 12:45, 14 May 2010
Line 15:		Line 15:
	This is the ''velocity'' of the aeroplane. The ''speed'' is the magnitude of the velocity,		This is the ''velocity'' of the aeroplane. The ''speed'' is the magnitude of the velocity,

-	<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>	+	<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>

-	~~The~~ horisontal speed is,	+	which is the answer to this part.
		+
		+
		+	It is worth noting that the horisontal speed is,
	<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44}=\sqrt{148}=12\textrm{.}2\ \text{m}{{\text{s}}^{-1}}</math>		<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44}=\sqrt{148}=12\textrm{.}2\ \text{m}{{\text{s}}^{-1}}</math>

Ian at 10:30, 20 April 2010

2010-04-20T10:30:12Z

←Older revision		Revision as of 10:30, 20 April 2010
Line 16:		Line 16:

	<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>		<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>
		+
		+	The horisontal speed is,
		+	<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44}=\sqrt{148}=12\textrm{.}2\ \text{m}{{\text{s}}^{-1}}</math>

Ian at 10:28, 20 April 2010

2010-04-20T10:28:08Z

←Older revision		Revision as of 10:28, 20 April 2010
Line 11:		Line 11:
	giving,		giving,

-	<math>\mathbf{v}=8\ \mathbf{i}+8 \ \mathbf{ j}+((0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}))\times 40 =8\textrm{.}4\mathbf{i}+8\textrm{.}8\mathbf{j}+12\mathbf{k}</math>	+	<math>\mathbf{v}=8\ \mathbf{i}+8 \ \mathbf{ j}+((0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}))\times 40 =8\textrm{.}4\mathbf{i}+8\textrm{.}8\mathbf{j}+4\mathbf{k}</math>

-	This is the ''velocity'' of the aeroplane. The ''speed'' is the magnitude of the velocity.	+	This is the ''velocity'' of the aeroplane. The ''speed'' is the magnitude of the velocity,
		+
		+	<math>\sqrt{{{8\textrm{.}4}^{2}}+{{8\textrm{.}8}^{2}}+{{4}^{2}}}=\sqrt{70\textrm{.}56+77\textrm{.}44+16}=\sqrt{164}=12\textrm{.}8\ \text{m}{{\text{s}}^{-1}}</math>

Ian at 10:16, 20 April 2010

2010-04-20T10:16:03Z

←Older revision		Revision as of 10:16, 20 April 2010
Line 8:		Line 8:

	<math>t=40 \ \text{s} </math>.		<math>t=40 \ \text{s} </math>.
		+
		+	giving,
		+
		+	<math>\mathbf{v}=8\ \mathbf{i}+8 \ \mathbf{ j}+((0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k}))\times 40 =8\textrm{.}4\mathbf{i}+8\textrm{.}8\mathbf{j}+12\mathbf{k}</math>
		+
		+	This is the ''velocity'' of the aeroplane. The ''speed'' is the magnitude of the velocity.

Ian: New page: We use the equation $\mathbf{v}=\mathbf{u}+\mathbf{a}t \$ where, $\mathbf{a}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}...$

2010-04-20T10:09:39Z

New page: We use the equation <math>\mathbf{v}=\mathbf{u}+\mathbf{a}t \ </math> where, <math>\mathbf{a}=(0\textrm{.}01\mathbf{i}+0\textrm{.}02\mathbf{j}+0\textrm{.}1\mathbf{k})\text{ m}{{\text{s}}...

New page