Solution 6.1
From Mechanics
(Difference between revisions)
(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...) |
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| - | <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> |
Revision as of 16:36, 4 March 2011
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 \displaystyle t-axis.
This latter method gives
\displaystyle \frac{1}{2}\left( 21+\left[ 16-7 \right] \right) \text{s}\times 9 \ \text{m}{{\text{s}}^{\text{-1}}}=135\ \text{m}
