Solution 18.1c - Revision history

Ian at 16:57, 27 March 2011

Ian — Sun, 27 Mar 2011 16:57:49 GMT

←Older revision		Revision as of 16:57, 27 March 2011
Line 2:		Line 2:

	<math>\begin{align}		<math>\begin{align}
-	& 2t-\frac{3{{t}^{2}}}{20}=0 \\	+	& 2t-\frac{{{t}^{2}}}{5}=0 \\
	& \text{which gives } \\		& \text{which gives } \\
-	& t=0\ \text{s }\ \text{or }\ t=~~\frac{40}{3}=13\textrm{.}3~~\ \text{s} \\	+	& t=0\ \text{s }\ \text{or }\ t=10\ \text{s} \\
	\end{align}</math>		\end{align}</math>

Ian at 15:30, 4 October 2010

Ian — Mon, 04 Oct 2010 15:30:40 GMT

←Older revision		Revision as of 15:30, 4 October 2010
Line 3:		Line 3:
	<math>\begin{align}		<math>\begin{align}
	& 2t-\frac{3{{t}^{2}}}{20}=0 \\		& 2t-\frac{3{{t}^{2}}}{20}=0 \\
-	& t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13.3\ \text{s} \\	+	& \text{which gives } \\
		+	& t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13\textrm{.}3\ \text{s} \\
	\end{align}</math>		\end{align}</math>

Ian: New page: Using the result of part b) with $a=0$ gives $\begin{align} & 2t-\frac{3{{t}^{2}}}{20}=0 \\ & t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13.3\ \text{s} \\ \end{align}<...$

Ian — Mon, 04 Oct 2010 15:29:18 GMT

New page: Using the result of part b) with <math>a=0</math> gives <math>\begin{align} & 2t-\frac{3{{t}^{2}}}{20}=0 \\ & t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13.3\ \text{s} \\ \end{align}<...

New page

Using the result of part b) with <math>a=0</math> gives

<math>\begin{align}
& 2t-\frac{3{{t}^{2}}}{20}=0 \\
& t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13.3\ \text{s} \\
\end{align}</math>

Solution 18.1c - Revision history

Ian at 16:57, 27 March 2011

Ian at 15:30, 4 October 2010

Ian: New page: Using the result of part b) with a=0 gives \begin{align} & 2t-\frac{3{{t}^{2}}}{20}=0 \\ & t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13.3\ \text{s} \\ \end{align}<...

Ian: New page: Using the result of part b) with $a=0$ gives $\begin{align} & 2t-\frac{3{{t}^{2}}}{20}=0 \\ & t=0\ \text{s }\ \text{or }\ t=\frac{40}{3}=13.3\ \text{s} \\ \end{align}<...$