The graph provided in your image shows the velocity $v_x$ of an object as a function of time $t$ in seconds (s), and the task is to determine the object's position at $t = 2$ seconds.

To find the position at $t = 2$ seconds, we'll need to integrate the velocity function from the graph. The graph represents a linear increase in velocity, suggesting that the velocity function $v_x(t)$ is linear, which can be written in the form $v_x(t) = mt + b$.

From the graph, at $t = 0$ seconds, $v_x = 0$ m/s, and at $t = 4$ seconds, $v_x = 12$ m/s. Therefore, the slope $m$ can be calculated as: $m = \frac{\Delta v}{\Delta t} = \frac{12 - 0}{4 - 0} = 3 \text{ m/s}^2$ Thus, $v_x(t) = 3t$.

The displacement $x$ can be found by integrating $v_x(t)$: $x(t) = \int v_x(t) \, dt = \int 3t \, dt = \frac{3}{2}t^2 + C$ At $t = 0$, assuming the initial position $x(0)$ is 0, we have: $x(0) = \frac{3}{2}(0)^2 + C = 0 \Rightarrow C = 0$ Thus, the position function is: $x(t) = \frac{3}{2}t^2$ At $t = 2$ seconds, the position $x$ is: $x(2) = \frac{3}{2}(2)^2 = 6 \text{ m}$

Therefore, the position of the object at $t = 2$ seconds is $6$ meters.

Would you like more details on this calculation or help with anything else?

Related Questions:

1. How would the position formula change if the initial velocity was not zero?
2. What would be the impact on the position if the acceleration was not constant?
3. How does the area under the velocity-time graph relate to displacement?
4. Can you calculate the average velocity between $t = 0$ and $t = 2$ seconds?
5. How would the position-time graph look based on the given velocity-time graph?

Tip: When dealing with linear motion where acceleration is constant, always remember that the area under the velocity-time graph gives you the displacement.