Let's break down the information from the velocity-time graph and the questions provided:

1. Graph Overview: The graph shows velocity (in m/s) on the y-axis and time (in seconds) on the x-axis. The particle starts at a velocity of 8 m/s at $t = 0$ and decelerates uniformly to $v = -2$ m/s at $t = 8$ seconds.

2. Acceleration of the Particle:

   • Acceleration is the slope of the velocity-time graph.
   • The change in velocity ($\Delta v$) is $-2 - 8 = -10 \, \text{m/s}$.
   • The time ($\Delta t$) over which this change happens is $8 \, \text{s}$.
   • Thus, the acceleration is: $a = \frac{\Delta v}{\Delta t} = \frac{-10}{8} = -1.25 \, \text{m/s}^2.$ It seems that the value in the box "0.5" is incorrect. The correct answer should be $-1.25 \, \text{m/s}^2$.

3. Total Distance Travelled:

   • The distance travelled is the area under the velocity-time graph.
   • The area is divided into two regions: a triangle above the x-axis and one below the x-axis.
   • Area of the first triangle (from 0 to 6 seconds, above the x-axis): $\text{Area} = \frac{1}{2} \times 6 \, \text{s} \times 8 \, \text{m/s} = 24 \, \text{m}.$
   • Area of the second triangle (from 6 to 8 seconds, below the x-axis): $\text{Area} = \frac{1}{2} \times 2 \, \text{s} \times 2 \, \text{m/s} = 2 \, \text{m}.$
   • Total distance travelled = $24 + 2 = 26 \, \text{m}$. The entered value of "16 m" seems incorrect. The correct distance should be $26 \, \text{m}$.

4. Displacement after 8 seconds:

   • The displacement is the net area under the graph (accounting for direction).
   • The area above the x-axis (from 0 to 6 seconds) is $+24 \, \text{m}$, while the area below the x-axis (from 6 to 8 seconds) is $-2 \, \text{m}$.
   • Thus, the total displacement is: $24 - 2 = 22 \, \text{m}.$

Would you like more details on any of these answers or need further clarification?

Related Questions:

1. How do you calculate average velocity from a velocity-time graph?
2. What is the significance of the area under a velocity-time graph?
3. How does acceleration relate to the slope of a velocity-time graph?
4. Can the displacement ever be greater than the total distance travelled?
5. How would the graph change if the particle had a constant positive acceleration?

Tip:

Always remember that in velocity-time graphs, the slope gives the acceleration, and the area under the graph gives the displacement.