Ian at 17:20, 10 March 2011

Ian — Thu, 10 Mar 2011 17:20:12 GMT

←Older revision		Revision as of 17:20, 10 March 2011
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	& 2(9\mathbf{i}+5\mathbf{j})+3(-3\mathbf{i}+4\mathbf{j})=(2+3)\mathbf{v} \\		& 2(9\mathbf{i}+5\mathbf{j})+3(-3\mathbf{i}+4\mathbf{j})=(2+3)\mathbf{v} \\
	& 18\mathbf{i}+10\mathbf{j}-9\mathbf{i}+12\mathbf{j}=5\mathbf{v} \\		& 18\mathbf{i}+10\mathbf{j}-9\mathbf{i}+12\mathbf{j}=5\mathbf{v} \\
-	& \mathbf{v}=\frac{9\mathbf{i}+22\mathbf{j}}{5}=(1\textrm{.}8\mathbf{i}+4\textrm{.}4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}} \\	+	& \mathbf{v}=\frac{\smash{9\mathbf{i}+22\mathbf{j}}}{5}=(1\textrm{.}8\mathbf{i}+4\textrm{.}4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}} \\
	\end{align}</math>		\end{align}</math>

Ian: New page: Using ${{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}$ and ${{\mathbf{u}}_{A}}={{\mathbf{u}}_{B}}$ ...

Ian — Thu, 23 Sep 2010 17:37:15 GMT

New page: Using <math>{{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}</math> and <math>{{\mathbf{u}}_{A}}={{\mathbf{u}}_{B}}</math>...

New page

Using

<math>{{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}</math>

and

<math>{{\mathbf{u}}_{A}}={{\mathbf{u}}_{B}}</math>

<math>\begin{align}
& 2(9\mathbf{i}+5\mathbf{j})+3(-3\mathbf{i}+4\mathbf{j})=(2+3)\mathbf{v} \\
& 18\mathbf{i}+10\mathbf{j}-9\mathbf{i}+12\mathbf{j}=5\mathbf{v} \\
& \mathbf{v}=\frac{9\mathbf{i}+22\mathbf{j}}{5}=(1\textrm{.}8\mathbf{i}+4\textrm{.}4\mathbf{j})\text{ m}{{\text{s}}^{\text{-1}}} \\
\end{align}</math>

Solution 16.6 - Revision history

Ian at 17:20, 10 March 2011

Ian: New page: Using {{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}} and {{\mathbf{u}}_{A}}={{\mathbf{u}}_{B}}...

Ian: New page: Using ${{m}_{A}}{{\mathbf{v}}_{A}}+{{m}_{B}}{{\mathbf{v}}_{B}}={{m}_{A}}{{\mathbf{u}}_{A}}+{{m}_{B}}{{\mathbf{u}}_{B}}$ and ${{\mathbf{u}}_{A}}={{\mathbf{u}}_{B}}$ ...