Ian at 17:55, 27 March 2011

2011-03-27T17:55:02Z

←Older revision		Revision as of 17:55, 27 March 2011
Line 8:		Line 8:
	& t=0,\ \mathbf{r}=400\mathbf{j}\ \Rightarrow \ {{d}_{1}}=0,\ {{d}_{2}}=400 \\		& t=0,\ \mathbf{r}=400\mathbf{j}\ \Rightarrow \ {{d}_{1}}=0,\ {{d}_{2}}=400 \\
	& \\		& \\
-	& \mathbf{r}=\left( 40t \right)\mathbf{i}+\left( -2t+400 \right)\mathbf{j}	+	& \mathbf{r}=\left( 40t \right)\mathbf{i}+\left( {-{t}^{2}}+400 \right)\mathbf{j}
	\end{align}</math>		\end{align}</math>

Ian: New page: The expression for the velocity $\mathbf{v}$ was obtained in part a), $\begin{align} & \mathbf{r}=\int{\mathbf{v}dt} \\ & \\ & =\left( 40t+{{d}_{1}} \right)\mathbf{i}+\...$

2010-10-11T16:52:56Z

New page: The expression for the velocity <math>\mathbf{v}</math> was obtained in part a), <math>\begin{align} & \mathbf{r}=\int{\mathbf{v}dt} \\ & \\ & =\left( 40t+{{d}_{1}} \right)\mathbf{i}+\...

New page

The expression for the velocity <math>\mathbf{v}</math> was obtained in part a),

<math>\begin{align}
& \mathbf{r}=\int{\mathbf{v}dt} \\
& \\
& =\left( 40t+{{d}_{1}} \right)\mathbf{i}+\left( -{{t}^{2}}+{{d}_{2}} \right)\mathbf{j} \\
& \\
& t=0,\ \mathbf{r}=400\mathbf{j}\ \Rightarrow \ {{d}_{1}}=0,\ {{d}_{2}}=400 \\
& \\
& \mathbf{r}=\left( 40t \right)\mathbf{i}+\left( -2t+400 \right)\mathbf{j}
\end{align}</math>

Solution 19.8b - Revision history

Ian at 17:55, 27 March 2011

Ian: New page: The expression for the velocity \mathbf{v} was obtained in part a), \begin{align} & \mathbf{r}=\int{\mathbf{v}dt} \\ & \\ & =\left( 40t+{{d}_{1}} \right)\mathbf{i}+\...

Ian: New page: The expression for the velocity $\mathbf{v}$ was obtained in part a), $\begin{align} & \mathbf{r}=\int{\mathbf{v}dt} \\ & \\ & =\left( 40t+{{d}_{1}} \right)\mathbf{i}+\...$