The graph you've shared appears to show velocity (v) on the vertical axis (in meters per second, m/s), and time (t) on the horizontal axis (in seconds, s). It's a piecewise linear graph that forms a triangle with a peak at $(3,6)$, indicating that the velocity increases from 0 m/s to 6 m/s in the first 3 seconds and then decreases back to 0 m/s at 6 seconds.

If this graph represents the velocity over time, we can calculate the displacement by finding the area under the curve (since displacement is the integral of velocity over time).

Steps:

1. Break the graph into two sections:

   • From $t = 0$ to $t = 3$ seconds: This is a triangle with a base of 3 seconds and a height of 6 m/s.
   • From $t = 3$ to $t = 6$ seconds: Another triangle with the same dimensions.

2. Calculate the area of each triangle:

   • Area of a triangle = $\frac{1}{2} \times \text{base} \times \text{height}$.

   • For both triangles, the base is 3 seconds and the height is 6 m/s, so the area of each triangle is: $\text{Area} = \frac{1}{2} \times 3 \times 6 = 9 \, \text{meters}.$

3. Total displacement: Since there are two identical triangles, the total displacement is: $\text{Total displacement} = 9 + 9 = 18 \, \text{meters}.$

Thus, the total displacement over the 6-second interval is 18 meters.

Would you like more details on this or have any questions? Here are five related questions:

1. How do you calculate velocity from displacement and time?
2. What is the significance of the slope of the line on a velocity-time graph?
3. How can we interpret negative velocities on a velocity-time graph?
4. Can we calculate acceleration from this graph?
5. What if the graph was not linear—how would the displacement be calculated?

Tip: When dealing with velocity-time graphs, the area under the curve always represents the displacement.