The problem asks to find the speed of the particle at $t = 6 \, \text{s}$, based on the provided acceleration-time graph.

Step-by-step solution:

1. Understanding the graph:

   • The graph shows acceleration $a(t)$ in $\text{m/s}^2$ versus time $t$ in seconds.
   • From $t = 0 \, \text{s}$ to $t = 4 \, \text{s}$, the acceleration is constant at $a = 2 \, \text{m/s}^2$.
   • From $t = 4 \, \text{s}$ to $t = 5 \, \text{s}$, the acceleration decreases linearly from $2 \, \text{m/s}^2$ to $0 \, \text{m/s}^2$.
   • After $t = 5 \, \text{s}$, the acceleration remains zero (constant velocity).

2. Using the velocity-acceleration relationship: The velocity of the particle is the integral of the acceleration over time: $v(t) = \int a(t) \, dt$ Since the particle starts from rest, $v(0) = 0$.

3. Calculating the velocity at different intervals:

   • From $t = 0 \, \text{s}$ to $t = 4 \, \text{s}$: The area under the acceleration-time curve is a rectangle. The change in velocity over this time interval is: $\Delta v_1 = a \times t = 2 \, \text{m/s}^2 \times 4 \, \text{s} = 8 \, \text{m/s}$ So, the velocity at $t = 4 \, \text{s}$ is $v(4) = 8 \, \text{m/s}$.

   • From $t = 4 \, \text{s}$ to $t = 5 \, \text{s}$: The area under the acceleration-time curve is a triangle with base 1 second and height 2 m/s². The change in velocity is: $\Delta v_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \, \text{s} \times 2 \, \text{m/s}^2 = 1 \, \text{m/s}$ So, the velocity at $t = 5 \, \text{s}$ is: $v(5) = v(4) + \Delta v_2 = 8 \, \text{m/s} + 1 \, \text{m/s} = 9 \, \text{m/s}$

   • From $t = 5 \, \text{s}$ to $t = 6 \, \text{s}$: The acceleration is zero, so the velocity remains constant: $v(6) = v(5) = 9 \, \text{m/s}$

Final answer:

The speed of the particle at $t = 6 \, \text{s}$ is $\boxed{9 \, \text{m/s}}$.

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How would the speed change if the acceleration continued to decrease after 5 seconds?
2. What would be the total displacement of the particle in the first 6 seconds?
3. How would the graph look if the particle decelerated to a stop after 5 seconds?
4. How can you calculate the average speed of the particle over the first 6 seconds?
5. What effect would a negative acceleration have on the speed of the particle?

Tip: The area under the acceleration-time graph gives the change in velocity, and the slope of a velocity-time graph gives the acceleration.