Ian at 09:36, 11 October 2010

Ian — Mon, 11 Oct 2010 09:36:03 GMT

←Older revision		Revision as of 09:36, 11 October 2010
Line 1:		Line 1:
		+	Using the result for the velocity obtained in part a)
		+
	<math>\begin{align}		<math>\begin{align}
	& s(50)=\int_{0}^{50}{v}dt \\		& s(50)=\int_{0}^{50}{v}dt \\

Ian: New page: $\begin{align} & s(50)=\int_{0}^{50}{v}dt \\ & \\ & =\int_{0}^{50}{\left( \frac{{{t}^{2}}}{50}+2 \right)dt} \\ & \\ & =\left[ \frac{{{t}^{3}}}{150}+2t \right]_{0}^{50} \\ & \\...$

Ian — Mon, 11 Oct 2010 09:17:53 GMT

New page: <math>\begin{align} & s(50)=\int_{0}^{50}{v}dt \\ & \\ & =\int_{0}^{50}{\left( \frac{{{t}^{2}}}{50}+2 \right)dt} \\ & \\ & =\left[ \frac{{{t}^{3}}}{150}+2t \right]_{0}^{50} \\ & \\...

New page

<math>\begin{align}
& s(50)=\int_{0}^{50}{v}dt \\
& \\
& =\int_{0}^{50}{\left( \frac{{{t}^{2}}}{50}+2 \right)dt} \\
& \\
& =\left[ \frac{{{t}^{3}}}{150}+2t \right]_{0}^{50} \\
& \\
& =\left( \frac{{{50}^{3}}}{150}+2\times 50 \right)-0 \\
& \\
& =933\text{ m}
\end{align}</math>

Solution 19.2b - Revision history

Ian at 09:36, 11 October 2010

Ian: New page: \begin{align} & s(50)=\int_{0}^{50}{v}dt \\ & \\ & =\int_{0}^{50}{\left( \frac{{{t}^{2}}}{50}+2 \right)dt} \\ & \\ & =\left[ \frac{{{t}^{3}}}{150}+2t \right]_{0}^{50} \\ & \\...

Ian: New page: $\begin{align} & s(50)=\int_{0}^{50}{v}dt \\ & \\ & =\int_{0}^{50}{\left( \frac{{{t}^{2}}}{50}+2 \right)dt} \\ & \\ & =\left[ \frac{{{t}^{3}}}{150}+2t \right]_{0}^{50} \\ & \\...$