- Level: International Baccalaureate
- Subject: Maths
- Word count: 1642
Stopping Distances
Extracts from this document...
Introduction
Stopping Distances
Portfolio
Mathematics SL Yr. 1
Period A
Speed (x) Vs. Thinking Distance (y)
Points plotted on the graph:
(32, 6)
(48, 9)
(64, 12)
(80, 15)
(96, 18)
(112, 21)
As you can see in the graph above all six points seem to line up in a fairly straight line. As speed increases, so does the thinking distance, always by a similar amount. This means that the speed to thinking distance ratio is constantly increasing as speed increases.
Functions:
Slope:
m= (y2 – y1) / (x2 – x1)
m= (12 – 6) / (64 – 32)
m= 6/32
m= 3/16
(32,6):
y – y1 = m(x – x1)
y – 6 = (3/16) (x – 32)
y – 6 = (3/16)x –6
y = (3/16)x
(64, 12):
y – y1 = m(x – x1)
y – 12 = (3/16) (x - 64)
y – 12 = (3/16)x – 12
y = (3/16)x
Percentage of error”
(64, 12)
y = (3/16)64
y = 12
(12 – 12) /12
≈ 0 %
The function, y = (3/16)x, matches the points, plotted on the graph, almost perfectly. You can see this by the calculated percentage of error which equals 0%. The function goes through every single point graphed. In the real life situation this means that the function represents the data recorded, while measuring
Middle
y = 0.0039(80)2 + 0.188(80) – 4.0096
y = 24.96 + 15.04 – 4.0096
y = 35.9904
(35.9904 – 38) / 38
≈ - 5.288 %
The Function, y = 0.0039x2 + 0.188x – 4.0096, used to represent these data points is a parabolic function. It is not quite as accurate as the function representing speed vs. braking distance. I used the first three data points to come up with an appropriate function, thus it goes right through these points. The last three points however lay slightly above the function, with percentage errors of ≈ -12.04 %, ≈ - 9.124 % and ≈ -5.288.This function has several other limitations aswell. First of all it only works for lower speeds rather than when the car travels at higher speeds. Also, it represents a parabolic function however the data points only includes the positive numbers, meaning half of a parabola.
y = ax2 + bx + c
1. (80, 38)
2. (96, 55)
3. (112, 75)
- 38 = (80)2a + (80)b + c
38 = 6400a + 80b +c
- 55 = (96)2a + (96)b + c
55 = 9216a + 96b + c
- 75 = (112)2a + (112)b + c
75 = 12544a + 112b + c
38 = 6400a + 80b + c
- 55 = 9216a + 96b + c
17 = 2816a + 16b
55 = 9216a + 96b + c
- 75 = 12544a + 112b + c
20 = 3328a + 16b
17 = 2816a + 16b
- 20 = 3328a + 16b
3 = 512a
a ≈ 0.005900
17 = 2816 (0.0059) + 16b
17 = 16.6144 + 16b
0.3856 = 16b
b ≈ 0.02410
38 = 6400 (0.0059) + 80 (0.0241) + c
38 = 37.76 + 1.928 + c
c ≈ -1.688
Conclusion
Percentage of error
(10, 2.5)
y = 0.0061(10)2 + 0.1709 (10) + 0.2827
y = 2.6017
(2.6017 – 2.5) / 2.5
≈ 4.068 %
(40, 17)
y = 0.0061(40)2 + 0.1709 (40) + 0.2827
y = 16.8787
(16.8787 – 17) / 17
≈ -2.376 %
(90, 65)
y = 0.0061(90)2 + 0.1709 (90) + 0.2827
y = 65.0739
(65.0739 – 65) / 65
≈ 0.1134%
(160, 180)
y = 0.0061(160)2 + 0.1709(160) + 0.2827
y = 183.7867
(183.7867 – 180) / 180
≈ 2.104 %
My model, y = 0.0061x2 + 0.1709x + 0.2827, does not fit the data of overall stopping distances for other speeds. It barely intersects the first point and is extremely off the next couple. Although the type of function (parabolic) fits the graph, several points
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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