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Solution 6.1

From Mechanics

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Current revision (12:06, 27 March 2011) (edit) (undo)
 
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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>

Current revision

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 trapezium as the graph in the figure defines a trapezium with the t-axis.

This latter method gives

2121+1679 ms-1=135 m