# Stopping Distances

Stopping Distances

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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 how long it took for a person to think to apply the brakes, at different speeds, perfectly. The fact that a linear function resembles this situation in a very accurate manner means that the speed to thinking distance ratio is constant.

Speed (x) Vs. Braking Distance (y)

Points plotted on the graph:

(32, 6)

(48, 14)

(64, 24)

(80, 38)

(96, 55)

(112, 75)

Different from the first graph displayed, the points on this graph are not all lined up in a straight line. Here, the shape occurs to be a parabola. This graph however only includes the non-negative numbers, meaning it is only half of a parabola

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

1. 14 = (48)2a + (48)b + c

14 = 2304a + 48b + c

1. 24 = (64)2a + (64)b + c

24 = 4096a + 64b + c

6 = 1024a + 32b + c

- 14 = 2304a + 48b + c

8 = 1280a + 16b

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