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.

Question

A two column table with 6 rows. The first column has, Time in seconds, and has the entries, 0, 1, 2, 3, 4. The second column, Velocity in meters per second, has the entries, 0, 9.8, 19.6, 29.4, 39.2.
The acceleration due to gravity is the constant of variation. What is the acceleration due to gravity of a falling object?

4.9 9.8 StartFraction m Over s squared EndFraction.
9.8 9.8 StartFraction m Over s squared EndFraction.
10.2 10.2 StartFraction m Over s squared EndFraction.
19.6 19.6 StartFraction m Over s squared EndFraction.

Answer 1

The relationship between velocity and time for a falling object is governed by the equation $v=gt$, where $v$ is the velocity, $g$ is the acceleration due to gravity, and $t$ is the time in seconds.

From the table you provided, we can see how the velocity changes with respect to time:

• At $t=0$, $v=0$
• At $t=1$, $v=9.8$
• At $t=2$, $v=19.6$
• At $t=3$, $v=29.4$
• At $t=4$, $v=39.2$

The increase in velocity per second is consistent, and each second corresponds to an increase of $9.8$ meters per second.

Since the velocity changes directly with time at a constant rate, the acceleration due to gravity $g$ can be determined by the increase in velocity over time:

• From $t=0$ to $t=1$ second: Increase of $9.8$ m/s
• From $t=1$ to $t=2$ seconds: Increase of $9.8$ m/s
• From $t=2$ to $t=3$ seconds: Increase of $9.8$ m/s
• From $t=3$ to $t=4$ seconds: Increase of $9.8$ m/s

Thus, the acceleration due to gravity $g$ is $9.8,{\text{m/s}}^2$.

Therefore, the correct answer is:

9.8 9.8 StartFraction m Over s squared EndFraction.