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\begin{tabular}{|l|l|}
\hline
Force & Acceleration \\
\hline
[tex]$0 \, \text{N}$[/tex] & [tex]$0 \, \text{m/s}^2$[/tex] \\
\hline
[tex]$150 \, \text{N}$[/tex] & [tex]$1 \, \text{m/s}^2$[/tex] \\
\hline
[tex]$200 \, \text{N}$[/tex] & [tex]$1.3 \, \text{m/s}^2$[/tex] \\
\hline
[tex]$250 \, \text{N}$[/tex] & [tex]$1.6 \, \text{m/s}^2$[/tex] \\
\hline
[tex]$350 \, \text{N}$[/tex] & [tex]$2.3 \, \text{m/s}^2$[/tex] \\
\hline
[tex]$400 \, \text{N}$[/tex] & [tex]$2.6 \, \text{m/s}^2$[/tex] \\
\hline
\end{tabular}

What is the relationship between force and acceleration as supported by the data table when mass is constant?

A. Force is inversely proportional to acceleration.
B. Force is directly proportional to acceleration.
C. Acceleration decreases when force increases.
D. Force is unrelated to acceleration and motion.


Sagot :

To determine the relationship between force and acceleration from the given data, we can perform a linear regression analysis. This involves finding the best-fit line that describes the relationship between the two variables.

Given Data Table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Force (N)} & \text{Acceleration (m/s}^2\text{)} \\ \hline 0 & 0 \\ \hline 150 & 1 \\ \hline 200 & 1.3 \\ \hline 250 & 1.6 \\ \hline 350 & 2.3 \\ \hline 400 & 2.6 \\ \hline \end{array} \][/tex]

1. Collect Data Points:
[tex]\[ (0, 0), (150, 1), (200, 1.3), (250, 1.6), (350, 2.3), (400, 2.6) \][/tex]

2. Perform Linear Regression:
We aim to fit a line [tex]\( y = mx + b \)[/tex] where [tex]\( y \)[/tex] is the acceleration, and [tex]\( x \)[/tex] is the force. Here, [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept.

3. Compute the Slope (m) and Intercept (b):
After performing the calculations, we find:
- Slope ([tex]\( m \)[/tex]): [tex]\( 0.006506024096385543 \)[/tex]
- Intercept ([tex]\( b \)[/tex]): [tex]\( 0.0028112449799198133 \)[/tex]

4. Interpret the Slope (m):
- The positive slope [tex]\( m \)[/tex] indicates that as the force increases, the acceleration also increases.
- Since [tex]\( m \)[/tex] is positive and significant, it shows a positive linear relationship between force and acceleration.

5. Determine the Relationship:
The positive slope confirms that force is directly proportional to acceleration. According to Newton's Second Law of Motion ([tex]\( F = ma \)[/tex]), for a constant mass [tex]\( m \)[/tex], the force [tex]\( F \)[/tex] is directly proportional to the acceleration [tex]\( a \)[/tex].

Thus, based on the computations and derived relationship:
- Force is directly proportional to acceleration.
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