- Find the slope and intercept of the best fit line, using =slope(distance, speed) and =intercept(distance, speed)
- Write the equation of the line.
- Fill in the predicted values.
| speed $( m / s )$ | distance $( m )$ |
|---|---|
| 7 | 1.51 |
| 11 | 1.31 |
| 16 | 3.93 |
| 15 | 4.33 |
| 20 | 2.97 |
| 25 | 7.00 |
| | Intercept (b) |
|---|---|
| | Slope (m) |
| | |
| Equation | |
| speed $( m / s )$ | distance $( m )$ |
|---|---|
| 3 | |
| 6 | |
| 9 | |
| 12 | |
| 15 | |
| 18 | |
| 21 | |
| 24 | |
| 27 | |
| 30 | |
| 33 | |
Format to two decimal places.
Answer (2)
Calculate the slope (m) of the best fit line using the formula: m = n ( ∑ x 2 ) − ( ∑ x ) 2 n ( ∑ x y ) − ( ∑ x ) ( ∑ y ) , resulting in approximately 0.3595.
Calculate the y-intercept (b) using the formula: b = n ( ∑ x 2 ) − ( ∑ x ) 2 ( ∑ y ) ( ∑ x 2 ) − ( ∑ x ) ( ∑ x y ) , resulting in approximately -2.289.
Write the equation of the best fit line as: distance = 0.3595 * speed - 2.289.
Calculate the predicted distance values for the given speeds (3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33) using the equation and round to two decimal places. The predicted distances are: -1.21, -0.13, 0.95, 2.03, 3.10, 4.18, 5.25, 6.34, 7.41, 8.49, 9.56.
d i s t an ce = 0.3595 ∗ s p ee d − 2.289
Explanation
Problem Analysis We are given a set of data points representing the speed of a vehicle and its corresponding stopping distance. Our goal is to find the line of best fit for this data, determine its equation, and then use this equation to predict the stopping distances for a new set of speeds. This involves calculating the slope and y-intercept of the best fit line.
Formulas for Slope and Intercept To find the slope (m) and y-intercept (b) of the best fit line, we'll use the following formulas:
Slope (m): m = n ( ∑ x 2 ) − ( ∑ x ) 2 n ( ∑ x y ) − ( ∑ x ) ( ∑ y )
Y-intercept (b): b = n ( ∑ x 2 ) − ( ∑ x ) 2 ( ∑ y ) ( ∑ x 2 ) − ( ∑ x ) ( ∑ x y )
Where:
n is the number of data points
x represents the speed values
y represents the distance values
Calculating Sums from Data First, let's calculate the necessary sums from the given data:
speed (x)
distance (y)
xy
x^2
7
1.51
10.57
49
11
1.31
14.41
121
16
3.93
62.88
256
15
4.33
64.95
225
20
2.97
59.40
400
25
7.00
175.00
625
---
---
---
---
\sum x = 94
\sum y = 20.05
\sum xy = 387.21
\sum x^2 = 1676
n = 6 (number of data points)
Calculating Slope and Intercept Now, we can plug these values into the formulas for the slope and y-intercept:
Slope (m): m = 6 ( 1676 ) − ( 94 ) 2 6 ( 387.21 ) − ( 94 ) ( 20.05 ) = 10056 − 8836 2323.26 − 1884.7 = 1220 438.56 ≈ 0.3595
Y-intercept (b): b = 6 ( 1676 ) − ( 94 ) 2 ( 20.05 ) ( 1676 ) − ( 94 ) ( 387.21 ) = 10056 − 8836 33603.8 − 36397.74 = 1220 − 2793.94 ≈ − 2.289
Equation of the Best Fit Line The equation of the best fit line is:
distance = m * speed + b distance = 0.3595 * speed - 2.289
Now, we can use this equation to predict the distance values for the given speeds.
Calculating Predicted Values Let's calculate the predicted distance values for the given speeds:
speed (m/s)
Predicted distance (m)
3
0.3595 * 3 - 2.289 = -1.21
6
0.3595 * 6 - 2.289 = -0.13
9
0.3595 * 9 - 2.289 = 0.95
12
0.3595 * 12 - 2.289 = 2.03
15
0.3595 * 15 - 2.289 = 3.10
18
0.3595 * 18 - 2.289 = 4.18
21
0.3595 * 21 - 2.289 = 5.25
24
0.3595 * 24 - 2.289 = 6.34
27
0.3595 * 27 - 2.289 = 7.41
30
0.3595 * 30 - 2.289 = 8.49
33
0.3595 * 33 - 2.289 = 9.56
Rounding to two decimal places:
speed (m/s)
Predicted distance (m)
3
-1.21
6
-0.13
9
0.95
12
2.03
15
3.10
18
4.18
21
5.25
24
6.34
27
7.41
30
8.49
33
9.56
Examples
Understanding the relationship between speed and stopping distance is crucial for road safety. For example, traffic engineers use this relationship to determine safe following distances and to design traffic signals that account for the time it takes vehicles to stop. Similarly, car manufacturers use this data to develop and test braking systems. By analyzing the best fit line, we can predict how much distance a car needs to stop at a given speed, which helps in preventing accidents and improving overall road safety.
The slope of the best fit line is approximately 0.3595, and the intercept is approximately -2.289, leading to the equation: distance = 0.3595 * speed - 2.289. Predicted distances for the given speeds rounded to two decimal places are provided in a tabulated format.
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