• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15

# 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)

1. 38 = (80)2a + (80)b + c

38 = 6400a + 80b +c

1. 55 = (96)2a + (96)b + c

55 = 9216a + 96b + c

1. 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

2 + 0.1709 + 0.2827, is very similar to the previous function, y = 0.006x2 – 0.0053x + 0.0256, representing speed vs. braking distance. Both are parabolic functions and intersect all points of the graph. IN addition, both have the same limitation which is the fact that the function represents an entire parabola, including both positive and negative points on the graph. The situation however only includes the positive numbers due to the real life situation in which it is physically impossible to have negative meters. This function is different to the function representing speed vs. thinking distance, y = (3/16)x, because it is a linear function representing one straight line. Also this function barely has any limitations when compared to the real life situation and the data recorded.

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.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## A logistic model

fish of a hydrolectric project during the first 30 years by means of the logistic function model Un+1 {19} considering a harvest of 8000 fish per annum. Initial population is 45,000 fish 45000 44000 43000 42000 41000 40000 39000 0 5 10 15 20 25 30 Year Figure 9.5.

2. ## Stopping distances portfolio. In this task, we may develop individual functions that model the ...

12 = 64m + c (96,18) 18 = 96m + c We consider it as a system of equations and solve it by substitution. Substitution c = 12 - 64m 18 = 96m + 12 - 64m 18-12 = 96m - 64m 6 = 32m m = 0.1875 So, if

1. ## portfolio Braking distance of cars

This is exponential function. The graph is almost through all of the point, but it is not absurdly correct. Now, check this function is appropriate. y= y=14 (braking distance), x=48(speed) 14=� 14= As a result, this function was not appropriate all value of table.

2. ## Stopping Distances

braking distance, there isn't a common difference and the braking distance rising exponentially creates a curve suggesting a parabola.

1. ## Creating a logistic model

number of fish will diminish. In this case, the number of fish has diminished to such an extreme extent that there are no fish left. So far we have been looking at examples where a population of fish has been left alone to cultivate.

2. ## The speed of Ada and Fay

Furthermore, I also set Ada's velocity as "u m/s" and Fay's velocity is "v m/s". Ada is running in a straight line along the harbor path using the velocity of 6 m/s and Fay is using 8m/s running towards Ada, which its direction will change due to Ada's position.

1. ## Modelling the amount of a drug in the bloodstre

Note that it is an exponential decay, it never touches the x-axis, although it gets arbitrarily close to it (thus the x-axis is a horizontal asymptote to the graph). Now that the function has been chosen, a question may have been asked on why this exponential function is formed as

2. ## Mathematic SL IA -Gold medal height (scored 16 out of 20)

Through the processes, I excluded 4 functions; sine, cosine, linear and quadratic functions. Since the data given has irregular patterns, I think that I can find best-fit model in polynomial functions. Thus, I am going to investigate the task from the cubic function.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to