Explore IDNLearn.com's extensive Q&A database and find the answers you're looking for. Our platform offers reliable and detailed answers, ensuring you have the information you need.
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]
[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]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.