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Revision as of 10:02, 7 April 2010 (edit) Ian (Talk | contribs) (New page: Using <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math> gives) ← Previous diff		Current revision (10:29, 7 April 2010) (edit) (undo) Ian (Talk | contribs)
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-	Using <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math> gives	+	Using <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as \ \</math>
		+	gives
		+
		+	<math>\begin{align}
		+	& {{15}^{2}}={{5}^{2}}+2\times a\times 500 \\
		+	& 200=1000a \\
		+	& a=0\textrm{.}2 \text{ m}{{\text{s}}^{\text{-2}}}\\
		+	& \\
		+	\end{align}</math>

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Current revision

Using \displaystyle {{v}^{\ 2}}={{u}^{\ 2}}+2as \ \ gives

\displaystyle \begin{align} & {{15}^{2}}={{5}^{2}}+2\times a\times 500 \\ & 200=1000a \\ & a=0\textrm{.}2 \text{ m}{{\text{s}}^{\text{-2}}}\\ & \\ \end{align}