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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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 | + | 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) | + | <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 \displaystyle t-axis.
This latter method gives
\displaystyle \frac{1}{2}\left( 21+\left[ 16-7 \right] \right) \times 9 \ \text{m}{{\text{s}}^{\text{-1}}}=135\ \text{m}
