the velocity of an object in meters per second varies directly with time in seconds since the object was…

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. the acceleration due to gravity is the constant of variation. what is the acceleration due to gravity of a falling object? velocity of a falling object\n| time (seconds) | velocity (meters/second) |\n| ---- | ---- |\n| 0 | 0 |\n| 1 | 9.8 |\n| 2 | 19.6 |\n| 3 | 29.4 |\n| 4 | 39.2 |\n4.9 \\frac{m}{s^{2}}\n9.8 \\frac{m}{s^{2}}\n10.2 \\frac{m}{s^{2}}\n19.6 \\frac{m}{s^{2}}

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. the acceleration due to gravity is the constant of variation. what is the acceleration due to gravity of a falling object? velocity of a falling object\n| time (seconds) | velocity (meters/second) |\n| ---- | ---- |\n| 0 | 0 |\n| 1 | 9.8 |\n| 2 | 19.6 |\n| 3 | 29.4 |\n| 4 | 39.2 |\n4.9 \\frac{m}{s^{2}}\n9.8 \\frac{m}{s^{2}}\n10.2 \\frac{m}{s^{2}}\n19.6 \\frac{m}{s^{2}}

Answer

Explanation:

Step1: Recall acceleration - velocity relationship

Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v$ is change in velocity and $\Delta t$ is change in time.

Step2: Select two - point data

Let's take the first two points $(t_1 = 0,v_1 = 0)$ and $(t_2=1,v_2 = 9.8)$.

Step3: Calculate acceleration

$a=\frac{v_2 - v_1}{t_2 - t_1}=\frac{9.8 - 0}{1 - 0}=9.8\frac{m}{s^2}$. We can also check with other pairs. For example, taking $(t_1 = 1,v_1 = 9.8)$ and $(t_2 = 2,v_2=19.6)$, $a=\frac{19.6 - 9.8}{2 - 1}=9.8\frac{m}{s^2}$.

Answer:

$9.8\frac{m}{s^2}$