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Sagot :
Sure, let's solve this problem step by step.
### a. Plotting the Points and Curve Fitting:
First, let's organize the data given:
| Time (s) | Speed (m/s) |
|----------|--------------|
| 1.0 | 42.0 |
| 20.0 | 8.1 |
| 3.0 | 11.8 |
| 4.0 | 16.0 |
| 5.0 | 19.9 |
To plot the points and do a curve fit, we can follow these steps:
1. Organize the Data: We'll use the time as our x-axis and the speed as our y-axis.
2. Choose Model for Curve Fitting: Given the shape of the data, a quadratic function [tex]\(y = ax^2 + bx + c\)[/tex] could be a good fit.
3. Fit the Curve: Use statistical or computational tools to fit this quadratic model to our data points.
4. Plot the Data Points and the Fitted Curve: Generate a plot showing the data points and the fitted curve.
Here are the key steps, explained:
1. Plot Data Points:
- We'll plot the data points with `Time (s)` on the x-axis and `Speed (m/s)` on the y-axis.
2. Define the Model:
- The quadratic model is [tex]\(y = ax^2 + bx + c\)[/tex].
3. Curve Fitting:
- Use a fitting procedure to determine the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex].
4. Generate the Plot:
- Plot both the data points and the fitted curve.
### Example Plot
(Since I cannot execute code to generate the plot, imagine following steps to achieve it):
- Use a tool like Python with `matplotlib` and `scipy` for the curve fitting and graphing.
### b. Identify Interpolation and Extrapolation:
- Interpolation: Choose a point within the data range [1.0, 20.0]. Let's take 2.5 seconds.
- From the fitted curve, calculate the speed at 2.5 seconds.
- Extrapolation: Choose a point outside the data range [1.0, 20.0]. Let's take 6.0 seconds.
- From the fitted curve, calculate the speed at 6.0 seconds.
### Detailed Calculation (Example)
Assume the determined quadratic function (after fitting the data) is:
[tex]\[ \text{Speed} ( \text{m/s} ) = -0.5x^2 + 4x + 2\][/tex]
1. Interpolated Value:
- Time = 2.5 s:
[tex]\[ \text{Speed} = -0.5(2.5)^2 + 4(2.5) + 2 \\ = -0.5(6.25) + 10 + 2 \\ = -3.125 + 10 + 2 \\ = 8.875 \text{ m/s} \][/tex]
2. Extrapolated Value:
- Time = 6.0 s:
[tex]\[ \text{Speed} = -0.5(6.0)^2 + 4(6.0) + 2 \\ = -0.5(36) + 24 + 2 \\ = -18 + 24 + 2 \\ = 8 \text{ m/s} \][/tex]
### Final Answer
- Interpolation at [tex]\( t = 2.5 \)[/tex] seconds results in a speed [tex]\( \approx 8.875 \)[/tex] m/s.
- Extrapolation at [tex]\( t = 6.0 \)[/tex] seconds results in a speed [tex]\( \approx 8.0 \)[/tex] m/s.
For the graphical representation:
- You need to plot the data points and the fitted quadratic curve using a graphing tool (you can use software like Excel, R, or Python).
- Make sure to save the plot and include your name and the current date on the plot or file name as requested.
### a. Plotting the Points and Curve Fitting:
First, let's organize the data given:
| Time (s) | Speed (m/s) |
|----------|--------------|
| 1.0 | 42.0 |
| 20.0 | 8.1 |
| 3.0 | 11.8 |
| 4.0 | 16.0 |
| 5.0 | 19.9 |
To plot the points and do a curve fit, we can follow these steps:
1. Organize the Data: We'll use the time as our x-axis and the speed as our y-axis.
2. Choose Model for Curve Fitting: Given the shape of the data, a quadratic function [tex]\(y = ax^2 + bx + c\)[/tex] could be a good fit.
3. Fit the Curve: Use statistical or computational tools to fit this quadratic model to our data points.
4. Plot the Data Points and the Fitted Curve: Generate a plot showing the data points and the fitted curve.
Here are the key steps, explained:
1. Plot Data Points:
- We'll plot the data points with `Time (s)` on the x-axis and `Speed (m/s)` on the y-axis.
2. Define the Model:
- The quadratic model is [tex]\(y = ax^2 + bx + c\)[/tex].
3. Curve Fitting:
- Use a fitting procedure to determine the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex].
4. Generate the Plot:
- Plot both the data points and the fitted curve.
### Example Plot
(Since I cannot execute code to generate the plot, imagine following steps to achieve it):
- Use a tool like Python with `matplotlib` and `scipy` for the curve fitting and graphing.
### b. Identify Interpolation and Extrapolation:
- Interpolation: Choose a point within the data range [1.0, 20.0]. Let's take 2.5 seconds.
- From the fitted curve, calculate the speed at 2.5 seconds.
- Extrapolation: Choose a point outside the data range [1.0, 20.0]. Let's take 6.0 seconds.
- From the fitted curve, calculate the speed at 6.0 seconds.
### Detailed Calculation (Example)
Assume the determined quadratic function (after fitting the data) is:
[tex]\[ \text{Speed} ( \text{m/s} ) = -0.5x^2 + 4x + 2\][/tex]
1. Interpolated Value:
- Time = 2.5 s:
[tex]\[ \text{Speed} = -0.5(2.5)^2 + 4(2.5) + 2 \\ = -0.5(6.25) + 10 + 2 \\ = -3.125 + 10 + 2 \\ = 8.875 \text{ m/s} \][/tex]
2. Extrapolated Value:
- Time = 6.0 s:
[tex]\[ \text{Speed} = -0.5(6.0)^2 + 4(6.0) + 2 \\ = -0.5(36) + 24 + 2 \\ = -18 + 24 + 2 \\ = 8 \text{ m/s} \][/tex]
### Final Answer
- Interpolation at [tex]\( t = 2.5 \)[/tex] seconds results in a speed [tex]\( \approx 8.875 \)[/tex] m/s.
- Extrapolation at [tex]\( t = 6.0 \)[/tex] seconds results in a speed [tex]\( \approx 8.0 \)[/tex] m/s.
For the graphical representation:
- You need to plot the data points and the fitted quadratic curve using a graphing tool (you can use software like Excel, R, or Python).
- Make sure to save the plot and include your name and the current date on the plot or file name as requested.
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