Answer :
To determine the acceleration due to gravity of a falling object, we need to understand how the velocity of the object changes with time. The relationship between time (t) and velocity (v) is given in the table provided.
[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]
Since the velocity of the falling object varies directly with time, the acceleration due to gravity is the rate of change of velocity with respect to time, which is constant.
1. To find the acceleration, we will use the data points. Let's consider two consecutive points for calculating acceleration.
2. Using the points (0 seconds, 0 m/s) and (1 second, 9.8 m/s):
[tex]\[ \text{Acceleration} = \frac{\text{Change in Velocity}}{\text{Change in Time}} \][/tex]
3. So we calculate:
[tex]\[ \text{Acceleration} = \frac{9.8 \, \text{m/s} - 0 \, \text{m/s}}{1 \, \text{second} - 0 \, \text{seconds}} = \frac{9.8 \, \text{m/s}}{1 \, \text{s}} = 9.8 \, \text{m/s}^2 \][/tex]
4. To confirm that this is indeed constant, we can check the next intervals in a similar manner:
- Between 1 second and 2 seconds:
[tex]\[ \text{Acceleration} = \frac{19.6 \, \text{m/s} - 9.8 \, \text{m/s}}{2 \, \text{seconds} - 1 \, \text{second}} = \frac{9.8 \, \text{m/s}}{1 \, \text{s}} = 9.8 \, \text{m/s}^2 \][/tex]
- Between 2 seconds and 3 seconds:
[tex]\[ \text{Acceleration} = \frac{29.4 \, \text{m/s} - 19.6 \, \text{m/s}}{3 \, \text{seconds} - 2 \, \text{seconds}} = \frac{9.8 \, \text{m/s}}{1 \, \text{s}} = 9.8 \, \text{m/s}^2 \][/tex]
- Between 3 seconds and 4 seconds:
[tex]\[ \text{Acceleration} = \frac{39.2 \, \text{m/s} - 29.4 \, \text{m/s}}{4 \, \text{seconds} - 3 \, \text{seconds}} = \frac{9.8 \, \text{m/s}}{1 \, \text{s}} = 9.8 \, \text{m/s}^2 \][/tex]
5. Since the acceleration is consistent across all time intervals and equals [tex]\(9.8 \frac{m}{s^2}\)[/tex], we conclude that:
The acceleration due to gravity of the falling object is [tex]\(9.8 \frac{m}{s^2}\)[/tex].
So, the correct answer is:
[tex]\[ \boxed{9.8 \frac{m}{s^2}} \][/tex]
[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]
Since the velocity of the falling object varies directly with time, the acceleration due to gravity is the rate of change of velocity with respect to time, which is constant.
1. To find the acceleration, we will use the data points. Let's consider two consecutive points for calculating acceleration.
2. Using the points (0 seconds, 0 m/s) and (1 second, 9.8 m/s):
[tex]\[ \text{Acceleration} = \frac{\text{Change in Velocity}}{\text{Change in Time}} \][/tex]
3. So we calculate:
[tex]\[ \text{Acceleration} = \frac{9.8 \, \text{m/s} - 0 \, \text{m/s}}{1 \, \text{second} - 0 \, \text{seconds}} = \frac{9.8 \, \text{m/s}}{1 \, \text{s}} = 9.8 \, \text{m/s}^2 \][/tex]
4. To confirm that this is indeed constant, we can check the next intervals in a similar manner:
- Between 1 second and 2 seconds:
[tex]\[ \text{Acceleration} = \frac{19.6 \, \text{m/s} - 9.8 \, \text{m/s}}{2 \, \text{seconds} - 1 \, \text{second}} = \frac{9.8 \, \text{m/s}}{1 \, \text{s}} = 9.8 \, \text{m/s}^2 \][/tex]
- Between 2 seconds and 3 seconds:
[tex]\[ \text{Acceleration} = \frac{29.4 \, \text{m/s} - 19.6 \, \text{m/s}}{3 \, \text{seconds} - 2 \, \text{seconds}} = \frac{9.8 \, \text{m/s}}{1 \, \text{s}} = 9.8 \, \text{m/s}^2 \][/tex]
- Between 3 seconds and 4 seconds:
[tex]\[ \text{Acceleration} = \frac{39.2 \, \text{m/s} - 29.4 \, \text{m/s}}{4 \, \text{seconds} - 3 \, \text{seconds}} = \frac{9.8 \, \text{m/s}}{1 \, \text{s}} = 9.8 \, \text{m/s}^2 \][/tex]
5. Since the acceleration is consistent across all time intervals and equals [tex]\(9.8 \frac{m}{s^2}\)[/tex], we conclude that:
The acceleration due to gravity of the falling object is [tex]\(9.8 \frac{m}{s^2}\)[/tex].
So, the correct answer is:
[tex]\[ \boxed{9.8 \frac{m}{s^2}} \][/tex]
