Stopping Distances

Functions:

y = ax2 + bx + c

1. (32, 6)

2. (48, 14)

3. (64, 24)

1. 6 = (32)2a + (32)b + c

6 = 1024a + 32b + c

2. 14 = (48)2a + (48)b + c

14 = 2304a + 48b + c

3. 24 = (64)2a + (64)b + c

24 = 4096a + 64b + c

     6 = 1024a + 32b + c

- 14 = 2304a + 48b + c

    8 = 1280a + 16b

   14 = 2304a + 48b + c

- 24 = 4096a + 64b + c

  10 = 1792a + 16b

    8 = 1280a + 16b

- 10 = 1792a + 16b

    2 = 512a

a ≈ 0.003900

8 = 1280 (0.0039) + 16b

8 = 4.992 + 16b

3.008 = 16b

b ≈ 0.1880

6 = 1024 (0.0039) + 32 (0.188) + c

6 = 3.9936 + 6.016 + c

c ≈ -4.0096

y = 0.0039x2 + 0.188x – 4.0096

Percentage of error

(112, 75)

y = 0.0039(112)2 + 0.188(112) – 4.0096

y = 48.9216 + 21.056 – 4.0096

y = 65.968

(65.968 – 75) / 75

≈ - 12.04 %

(96, 55)

y = 0.0039(96)2 + 0.188(96) – 4.0096

y = 35.9424 + 18.048 – 4.0086

y = 49.9818

(49.9818 – 55) / 55

≈ - 9.124 %

(80, 38)

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

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

55 = 9216a + 96b + c

3. 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

y = 0.0059x2 + 0.0241x – 1.688

The points used to find a new function were the last three points. Different from the previous function, this one works only for cars at higher speed, considering it goes right through the last couple of point however is slightly below the first two.

Percentage of error

(32, 6)

y = 0.0059(32)2 + 0.0241(32) – 1.688

y = 6.0416 + 0.7712 – 1.688

y = 5.1248

(5.1248 – 6) / 6

≈ -14.59%

1. (32, 6)

2. (80, 38)

3. (112, 75)

1. 6 = 1024a + 32b + c

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

38 = 6400a + 80b + c

3. 75 = 12544a + 112b + c

    6 = 1024a + 32b + c

- 38 = 6400a + 80b + c

   32 = 5376a + 48b

  38 = 6400a + 80b + c

- 75 = 12544a + 112b + c

37 = 6144a + 32b

(32 = 5376a + 48b) x 2

64 = 10752a + 96b

(37 = 6144a + 32b) x 3

111 = 18432a + 96b

    64 = 10752a + 96b

- 111 = 18432a + 96b

     47 = 7680a

a ≈ 0.006

32 = 5376 (0.006) + 48b

32 = 32.256 + 48b

-0.256 = 48b

b ≈ -0.0053

6 = 1024 (0.006) + 32 (-0.0053) + c

6 = 6.144 – 0.1696 + c

6 = 5.9744 + c

c ≈ 0.0256

y = 0.006x2 – 0.0053x + 0.0256

This parabolic function is combining the previous two functions, y = 0.0059x2 + 0.0241x – 1.688 and y = 0.0039x2 + 0.188x – 4.0096. It seems to fit quite well for almost all functions however still passes through the center of some points whereas others it barely touches the dot. A limitation for this graph, as for the previous two, is that it only includes positive numbers, meaning only half the parabola. This is because in the real life situation it would be impossible for the distance to be of negative amount. The calculated function however represents both positive and negative, creating a small restraint while representing speed vs. braking distance.

As you can see, as the speed of the car increases, so does the braking distance. These two variables however don’t increase at a constant ratio. This is apparent by studying the points on the graph and determining that they do not form a straight line however turn slightly upwards as they continue.

1. (32, 12)

2. (80, 53)

3. (112, 96)

1. 12 = (32)2a + (32)b + c

12 = 1024a + 32b + c

2. 53 = (80)2a + (80)b + c

53 = 6400a + 80b + c

3. 96 = (112)2a + (112)b + c

96 = 12544a + 112b + c

  12 = 1024a + 32b + c

- 53 = 6400a + 80b + c

   41 = 5376a + 48b

  53 = 6400a + 80b + c

- 96 = 12544a + 112b + c

   43 = 6144a + 32b

(41 = 5376a + 48b) x 2

82 = 10752a + 96b

(43 = 6144a + 32b) x 3

129 = 18432a + 96b

    82 = 10752a + 96b

- 129 = 18432a + 96b

     47 = 7680a

a ≈ 0.0061

41 = 5376 (0.0061) + 48b

b ≈ 0.1709

12 = 1024 (0.0061) + 32 (0.170967) + c

12 = 6.2464 + 5.470944 + c

c ≈ 0.2826

y = 0.0061x2 + 0.1709x + 0.2827

This is also a parabolic function. It works for all of the points and is thus a very good function to represent the speed vs. stopping distance. This function, y = 0.0061x2 + 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