Ian at 17:40, 27 March 2011

Ian — Sun, 27 Mar 2011 17:40:07 GMT

←Older revision		Revision as of 17:40, 27 March 2011
Line 2:		Line 2:

	<math>\begin{align}		<math>\begin{align}
-	& s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{~~128~~} \right)}dt \\	+	& s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{320} \right)}dt \\
	& \\		& \\
-	& =\left[ 20t-\frac{{{t}^{3}}}{~~128~~} \right]_{0}^{40} \\	+	& =\left[ 20t-\frac{{{t}^{3}}}{960} \right]_{0}^{40} \\
	& \\		& \\
-	& =\left( 20\times 40-\frac{{{40}^{3}}}{~~128~~} \right)-0 \\	+	& =\left( 20\times 40-\frac{{{40}^{3}}}{960} \right)-0 \\
	& \\		& \\
-	& =~~300~~\text{ m}	+	& =733\text{ m}
	\end{align}</math>		\end{align}</math>

Ian: New page: Using the expression for $v$ obtained in part a), $\begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{12...$

Ian — Mon, 11 Oct 2010 15:50:17 GMT

New page: Using the expression for <math>v</math> obtained in part a), <math>\begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{12...

New page

Using the expression for <math>v</math> obtained in part a),

<math>\begin{align}
& s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\
& \\
& =\left[ 20t-\frac{{{t}^{3}}}{128} \right]_{0}^{40} \\
& \\
& =\left( 20\times 40-\frac{{{40}^{3}}}{128} \right)-0 \\
& \\
& =300\text{ m}
\end{align}</math>

Solution 19.5b - Revision history

Ian at 17:40, 27 March 2011

Ian: New page: Using the expression for v obtained in part a), \begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{12...

Ian: New page: Using the expression for $v$ obtained in part a), $\begin{align} & s=\int_{0}^{40}{\left( 20-\frac{3{{t}^{2}}}{128} \right)}dt \\ & \\ & =\left[ 20t-\frac{{{t}^{3}}}{12...$