Solution 6.1 - Revision history

Ian at 12:06, 27 March 2011

Ian — Sun, 27 Mar 2011 12:06:04 GMT

←Older revision		Revision as of 12:06, 27 March 2011
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	This area can be calculated either by breaking up the area into three parts, two triangles and a rectangle		This area can be calculated either by breaking up the area into three parts, two triangles and a rectangle
-	or by using the formula for the area of a ~~trapedium~~ as the graph in the figure defines a trapezium with the <math>t</math>-axis.	+	or by using the formula for the area of a trapezium as the graph in the figure defines a trapezium with the <math>t</math>-axis.

	This latter method gives		This latter method gives

-	<math>\frac{1}{2}\left( 21+\left[ 16-7 \right] \right) ~~\text{s}~~\times 9 \ \text{m}{{\text{s}}^{\text{-1}}}=135\ \text{m}</math>	+	<math>\frac{1}{2}\left( 21+\left[ 16-7 \right] \right) \times 9 \ \text{m}{{\text{s}}^{\text{-1}}}=135\ \text{m}</math>

Ian at 16:36, 4 March 2011

Ian — Fri, 04 Mar 2011 16:36:38 GMT

←Older revision		Revision as of 16:36, 4 March 2011
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	This latter method gives		This latter method gives

-	<math>\frac{1}{2}\left( 21+\left[ 16-7 \right] \right) \text{s}\times 9 \text{m}{{\text{s}}^{\text{-1}}}=135\ \text{m}</math>	+	<math>\frac{1}{2}\left( 21+\left[ 16-7 \right] \right) \text{s}\times 9 \ \text{m}{{\text{s}}^{\text{-1}}}=135\ \text{m}</math>

Ian: New page: The area under the velocity-time graph gives the distance travelled. This area can be calculated either by breaking up the area into three parts, two triangles and a rectangle or by using...

Ian — Wed, 07 Apr 2010 08:43:28 GMT

New page: The area under the velocity-time graph gives the distance travelled. This area can be calculated either by breaking up the area into three parts, two triangles and a rectangle or by using...

New page

The area under the velocity-time graph gives the distance travelled.

This area can be calculated either by breaking up the area into three parts, two triangles and a rectangle
or by using the formula for the area of a trapedium as the graph in the figure defines a trapezium with the <math>t</math>-axis.

This latter method gives

<math>\frac{1}{2}\left( 21+\left[ 16-7 \right] \right) \text{s}\times 9 \text{m}{{\text{s}}^{\text{-1}}}=135\ \text{m}</math>