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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?

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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.

A particle starts from rest and moves in a straight line. Its acceleration-time graph is shown. Find the speed of the particle at t = 6 s.

This problem involves determining the speed of a particle at t = 6 s using an acceleration-time graph. Learn how to integrate acceleration over time to find velocity, including step-by-step calculations of velocity changes at different time intervals.

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