To determine the acceleration during the time interval from 0 to 2 seconds, we can use the following formula from kinematics:

$a = \frac{\Delta v}{\Delta t}$

where:

• $a$ is the acceleration,
• $\Delta v$ is the change in velocity,
• $\Delta t$ is the change in time.

From the table, at $t = 0$ seconds, the velocity is $0 \, \text{m/s}$, and at $t = 2$ seconds, the velocity is $1 \, \text{m/s}$.

Now, let's calculate the acceleration:

$a = \frac{1 \, \text{m/s} - 0 \, \text{m/s}}{2 \, \text{s} - 0 \, \text{s}} = \frac{1}{2} = 0.5 \, \text{m/s}^2$

So, the acceleration over the time interval from 0 to 2 seconds is $0.5 \, \text{m/s}^2$.

Would you like further details or have any questions?

Here are 5 related questions:

1. What is the acceleration for the interval from 2 to 3 seconds?
2. How does constant acceleration affect the velocity of an object over time?
3. What kinematic equations can be used to describe motion with constant acceleration?
4. How would the position change if the acceleration were higher during the interval?
5. How does the change in velocity affect the total distance traveled?

Tip: In problems with constant acceleration, the slope of a velocity vs. time graph gives you the acceleration.