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The velocity of an object in meters per second varies directly with time in seconds since the object was dropped, as represented by the table below.

| Time (seconds) | Velocity (meters/second) |
|----------------|--------------------------|
| 0 | 0 |
| 1 | 9.8 |
| 2 | 19.6 |
| 3 | 29.4 |
| 4 | 39.2 |

The acceleration due to gravity is the constant of variation. What is the acceleration due to gravity of a falling object?

A. [tex]\(4.9 \frac{m}{s^2}\)[/tex]
B. [tex]\(9.8 \frac{m}{s^2}\)[/tex]
C. [tex]\(10.2 \frac{m}{s^2}\)[/tex]
D. [tex]\(19.6 \frac{m}{s^2}\)[/tex]


Sagot :

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]