<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/css" href="http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/skins/common/feed.css?97"?>
<feed xmlns="http://www.w3.org/2005/Atom" xml:lang="en">
		<id>http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?action=history&amp;feed=atom&amp;title=Solution_6.2a</id>
		<title>Solution 6.2a - Revision history</title>
		<link rel="self" type="application/atom+xml" href="http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?action=history&amp;feed=atom&amp;title=Solution_6.2a"/>
		<link rel="alternate" type="text/html" href="http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?title=Solution_6.2a&amp;action=history"/>
		<updated>2026-05-28T08:46:44Z</updated>
		<subtitle>Revision history for this page on the wiki</subtitle>
		<generator>MediaWiki 1.11.1</generator>

	<entry>
		<id>http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?title=Solution_6.2a&amp;diff=982&amp;oldid=prev</id>
		<title>Ian: New page: The area under the velocity-time graph gives the distance travelled.  This area can be calculated using the formula for the area of a trapedium as the graph in the figure defines a trapezi...</title>
		<link rel="alternate" type="text/html" href="http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php?title=Solution_6.2a&amp;diff=982&amp;oldid=prev"/>
				<updated>2010-04-07T08:49:40Z</updated>
		
		<summary type="html">&lt;p&gt;New page: The area under the velocity-time graph gives the distance travelled.  This area can be calculated using the formula for the area of a trapedium as the graph in the figure defines a trapezi...&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;The area under the velocity-time graph gives the distance travelled.&lt;br /&gt;
&lt;br /&gt;
This area can be calculated using the formula for the area of a trapedium as the graph in the figure defines a trapezium with the &amp;lt;math&amp;gt;t&amp;lt;/math&amp;gt;-axis.&lt;br /&gt;
&lt;br /&gt;
This gives&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;\frac{1}{2}\left( 180+\left[ 120-40 \right] \right) \text{s}\times 24 \text{m}{{\text{s}}^{\text{-1}}}=3120\ \text{m}&amp;lt;/math&amp;gt;&lt;/div&gt;</summary>
		<author><name>Ian</name></author>	</entry>

	</feed>