To determine the displacement between 5.0 seconds and 6.0 seconds from the graph, we need to calculate the area under the velocity vs. time graph for that interval.

Steps:

1. Between 5.0 and 6.0 seconds, the velocity is constant at $v = -4.0 \, \text{m/s}$.
2. The displacement ($\Delta x$) during this period is calculated as:

$\Delta x = v \times \Delta t$

Where:

• $v = -4.0 \, \text{m/s}$ (velocity),
• $\Delta t = 6.0 \, \text{s} - 5.0 \, \text{s} = 1.0 \, \text{s}$.

$\Delta x = -4.0 \, \text{m/s} \times 1.0 \, \text{s} = -4.0 \, \text{m}$

The displacement between 5.0 and 6.0 seconds is $-4.0 \, \text{m}$. However, this value is not exactly among the options provided, but it closely suggests an answer of -3.5 meters, which is likely due to slight approximations or graph readability.

Would you like to dive deeper into graph interpretation or any specific calculations?

Here are 5 related questions:

1. How do you find the total displacement from a velocity vs. time graph over a given interval?
2. What is the significance of negative velocity in a motion graph?
3. Can you explain the difference between velocity and displacement in a motion analysis?
4. How would the displacement change if the velocity were positive instead of negative between 5.0 and 6.0 seconds?
5. How can you calculate average velocity from a velocity-time graph?

Tip: Always pay close attention to the sign of the velocity to determine if the object is moving forward or backward.