Equation B
The equation we are given to draw the graph is B: V = 0.5t where 0 ≤ t ≤ 9. This graph is a little more complicated than the last, so again to help myself I will draw a table of results:
After plotting the graph I discovered the shape of it to be a triangle. Using my theory that the area under a velocity/time graph equals the distance travelled, I was able to find the distance, I just had to remember the formula for working out the area of triangles. The formula for working out the area of triangles is ½(Base X Height) = Area. Using the lengths of the triangles, I was able to write ½(9 X 4.5) = Area, so the area is 20.25. As the area is the same as the distance travelled, this means the distance travelled is 20.25 metres.
As there is a positive gradient in this graph, this indicates acceleration. To work out acceleration we must firstly find the change in velocity, then divide it by the time taken. We can also work out what acceleration is measured in by putting the values into the formula – m/s divided by s = units of acceleration. If we carry out this formula, we will get m/s2 = Units of acceleration. By using the formula change in velocity divided by time taken = acceleration we can write 4.5 divided by 9 = Acceleration. So the acceleration for this graph is 0.5m/s2.
Equation C
For this graph we are given the equation C: V = t + 2 where 0 ≤ t ≤ 7. To help myself again I will draw a table of results:
After plotting the graph I discovered the graph was in the shape of a trapezium. As I know the formula for working out the area of a trapezium, I could find the area of this trapezium, and so the distance travelled. The formula for working out the area of trapeziums is ½(a+b)h where a & b are the parallel sides and h is the base length. After inserting the values into the formula I came up with ½(2+9) X 7 = 38.5. Because the area under a graph is equal to the distance travelled this means the distance travelled is also 38.5 metres.
In the graph there is a positive gradient, meaning there is acceleration. As the gradient is a straight line this indicates the acceleration is constant. By using the formula for working out acceleration we can write 9 – 2 = 7, which is the change in velocity. We can then write 7 divided by 7 = acceleration. So the acceleration is 1m/s2.
Equation D
This graph is similar to the last, in that it has the shape of a trapezium. We are given that D: V = 2t where 0 < t < 3 and V = 6 where 3 ≤ t ≤ 5. The table of results for this graph looks like this:
The formula I will need for this graph is exactly the same as the last graph – ½(a+b)h = area. The only difference with this graph is that the parallel lines are in a different position to the last. Again, by inserting the values I can write: ½(2+5) X 6 which is 21 metres. There is acceleration in this graph, but only for the first 3 seconds, after that, the speed is constant. I can work out the acceleration of the first part of the graph by writing 6 divided by 3 = 2m/s2.
Equation E
We are given that E: V = 8t – t where 0 ≤ t ≤ 8. This graph posed a few problems at first, as it was a curve. The results table looks like this:
After plotting the points I struggled to find the area under the curve until I remembered something we had been taught in maths – the trapezium rule. It is very hard to get the exact area under a curve with the trapezium rule, which is why it is called an estimate when we find the area. Here is a diagram of the trapezium rule being used in action:
Estimated Area = ½( y0 + 2y1 + 2y2 + 2y3 + 2y4 + 2y5)h
So for this graph, if we inserted the values, it would read: ½(0 + 2X12 + 2X16 + 2X12 + 0) X 2 = 80 Metres.
The only problem with this method is that every trapezium under the graph misses out a chunk of the graph. Here is a diagram showing what I’ve just mentioned:
There is a way to decrease the amount of graph left out by the trapeziums, and that is to use more trapeziums under the graph. I will draw the same graph, only this time I will use eight trapeziums instead of four.
If we inserted the values for this graph into the formula for the trapezium rule, it would read: ½(0 + 2X7 + 2X12 + 2X15 + 2X16 + 2X15 + 2X12 + 2X7 + 0) X 1 = 84 Metres.
Even though this was fairly accurate, I still wanted to get as close to the real value as possible. After searching through my maths textbook I found something called the mid-ordinate rule. The mid-ordinate rule works like the trapezium rule, but with rectangles. Here’s a small diagram of what it looks like:
The mid-ordinate rule is a lot more accurate than the trapezium rule, here is a magnified version of where you lose and gain with the mid-ordinate rule:
To properly plot the mid-ordinate rule, we must get a new table of results:
This will tell me the exact height of each rectangle, and where to plot them.
The formula for the mid-ordinate rule is (y1+y2+y3+y4+y5…)h where y is the height of the rectangle and h is the width. I can insert the values into this formula to work out the area of graph E: (3.75 + 9.75 + 13.75 + 15.75 + 15.75 + 13.75 + 9.75 + 3.75) X 1
Or more simply: 2(3.75 + 9.75 + 13.75 + 15.75) = Area. After calculating this, the area came to 86, which means the distance travelled was 86 metres.
As the acceleration is constantly changing in this graph, it is impossible to give an overall acceleration. Instead, it is only possible to find the acceleration at a certain given point. This can be done by drawing a tangent to the graph, and then making a right angled triangle out the tangent. This method will only work if the lengths either side of the tangent are equal. Also, it is only an estimate as there is always room for error.
On this graph I picked my point at two seconds, and then drew a tangent and a right-angled triangle. To get the acceleration we must divide the vertical side by the horizontal side, which in this case is: 2.8 divided by 1.4, which gives me 2m/s2
Equation F
We are given that F: V = t3. The results table looks like this:
After using the mid-ordinate rule last time and finding it more accurate, I decided to use it this time over the trapezium rule. This meant that I would have to draw another table of results to plot the rectangles:
After plotting the graph I used the formula from the last graph: (y1+y2+y3+y4+y5…)h and inserted the values into it. This gave me: 0.5(0.156...+ 0.4211…+ 1.9531… + 5.359… + 11.39… + 20.797…) = 19.97 metres.
Again as this graph is a curve, I cannot find the acceleration for the whole graph, only at one given point. I drew my tangent and right angle at 2.5 seconds, and then noted down the length of the vertical side and the horizontal side. This time the equation was 3.6 divided by 1.6 = 2.25m/s2
Conclusion
After testing each method it became apparent to me that the best method for finding area under a curve was the mid-ordinate rule because it was the most accurate out of the methods I had used. However, I wanted to be even more precise as the mid-ordinate rule was only an estimate, and so I started researching an A-level technique called intergration.
