To calculate the average acceleration ($a_{av,x}$) between $t = 7.00 \, \text{s}$ and $t = 11.0 \, \text{s}$, we use the formula for average acceleration:

$a_{av,x} = \frac{\Delta v_x}{\Delta t}$

Where:

• $\Delta v_x$ is the change in velocity in the x-direction.
• $\Delta t$ is the change in time.

Step 1: Find $v_x$ at $t = 7.00 \, \text{s}$

From the graph, at $t = 7.00 \, \text{s}$, the velocity $v_x$ is 5 m/s.

Step 2: Find $v_x$ at $t = 11.0 \, \text{s}$

At $t = 11.0 \, \text{s}$, the velocity $v_x$ is 15 m/s.

Step 3: Calculate $\Delta v_x$

$\Delta v_x = v_x(11.0 \, \text{s}) - v_x(7.00 \, \text{s}) = 15 \, \text{m/s} - 5 \, \text{m/s} = 10 \, \text{m/s}$

Step 4: Calculate $\Delta t$

$\Delta t = 11.0 \, \text{s} - 7.00 \, \text{s} = 4.00 \, \text{s}$

Step 5: Calculate $a_{av,x}$

$a_{av,x} = \frac{10 \, \text{m/s}}{4.00 \, \text{s}} = 2.5 \, \text{m/s}^2$

Thus, the average acceleration $a_{av,x}$ between $t = 7.00 \, \text{s}$ and $t = 11.0 \, \text{s}$ is $2.5 \, \text{m/s}^2$.

Do you have any questions or would you like further details on any part of the solution?

Here are five related questions to extend your understanding:

1. How would you calculate the average velocity in the same time interval?
2. What is the significance of the slope of the $v_x(t)$ graph?
3. How would you determine the total distance traveled between $t = 7.00 \, \text{s}$ and $t = 11.0 \, \text{s}$?
4. What would the graph of acceleration versus time look like based on this velocity graph?
5. How does constant velocity relate to zero acceleration?

Tip: The area under the velocity-time graph gives the displacement over a time interval.