Area under straight-line graphs
To calculate areas under graphs it is essential you remember your formulae. In particular the ones you will need are area of triangles, area of a rectangle and area of a trapezium.
Question
Work out the area under the graph:
\(Area = \frac {1} {2} \times {b} \times {h}\)
\(Area = \frac {1} {2} \times {2000} \times {400}\)
\(Area = {400,000}\)
Area under velocity-time graphs
In particular, if you find the area under Velocity-Time Graphs, you will need to calculate the distance travelled during the journey.
If you only want the distance up to a certain time, you find the area under the graph up until that time.
A cyclist leaves his house and accelerates uniformly until he reaches a speed of 8 m/s, he then travels at constant speed until he realises he has forgotten his money so slows down to a stop. Find the distance travelled when:
- accelerating
Where the slope is going in an upwards direction is where the cyclist is accelerating. We need to find the area of the triangle which ends at 4s.
\(Area = \frac {1} {2} \times {b} \times {h}\)
\(Area = \frac {1} {2} \times {4} \times {8}\)
\(Area = {16}\)
\(Area = {16}~{m}\)
- travelling at constant speed
The cyclist travels at a constant speed between 4 and 7 seconds. If we look at the shape under the graph between these times it is a rectangle.
\(Area = {l} \times {w}\)
\(Area = {3} \times {8}\)
\(Area = {24}\)
\(Area = {24}~{m}\)
- decelerating
The cyclist decelerates at the last stage of the journey, which again forms a triangle.
\(Area = \frac {1} {2} \times {b} \times {h}\)
\(Area = \frac {1} {2} \times {3} \times {8}\)
\(Area = {12}\)
\(Area = {12}~{m}\)
If we were asked to find the distance travelled through the whole journey, there are two methods.
Either the total of the distances at each stage of the journey.
Total distance = 16 + 24 + 12 = 52 m
In this case however the whole graph is in the shape of a trapezium. So we could work out the area of the trapezium.
\(A = \frac {1} {2} ({a}~+~{b}){h}\)
\(A = \frac {1} {2} ({10}~+~{3}){8}\)
\(A = {52}\)
Total distance = 52 m
