To determine the acceleration due to gravity, we will utilize the relationship between velocity and time, which is given as a direct variation. This can be expressed with the formula:
[tex]\[ v = a \cdot t \][/tex]
where [tex]\( v \)[/tex] is velocity, [tex]\( a \)[/tex] is acceleration, and [tex]\( t \)[/tex] is time.
We can use any two points from the data table provided to calculate the acceleration. Here’s the data given:
[tex]\[
\begin{array}{|c|c|}
\hline
\text{Time (seconds)} & \text{Velocity (meters/second)} \\
\hline
0 & 0 \\
\hline
1 & 9.8 \\
\hline
2 & 19.6 \\
\hline
3 & 29.4 \\
\hline
4 & 39.2 \\
\hline
\end{array}
\][/tex]
We will choose any two consecutive data points to calculate the acceleration. Let's use the points at [tex]\( t = 1 \)[/tex] second and [tex]\( t = 2 \)[/tex] seconds.
For [tex]\( t = 1 \)[/tex] second:
- [tex]\( t_1 = 1 \)[/tex]
- [tex]\( v_1 = 9.8 \)[/tex]
For [tex]\( t = 2 \)[/tex] seconds:
- [tex]\( t_2 = 2 \)[/tex]
- [tex]\( v_2 = 19.6 \)[/tex]
Acceleration [tex]\( a \)[/tex] can be calculated using the formula:
[tex]\[ a = \frac{v_2 - v_1}{t_2 - t_1} \][/tex]
Plugging in the values:
[tex]\[ a = \frac{19.6 - 9.8}{2 - 1} \][/tex]
[tex]\[ a = \frac{9.8}{1} \][/tex]
[tex]\[ a = 9.8 \][/tex]
Therefore, the acceleration due to gravity of a falling object, which is the constant of variation, is:
[tex]\[ \boxed{9.8 \frac{m}{s^2}} \][/tex]
