https://r2.erweima.ai/mathbot/image/17266010010335dn0ui9x.jpg

Let's analyze the problem step by step based on the graph provided.

### Problem Statement:
A particle starts from rest and moves according to the velocity-time graph shown. We are tasked with determining:
- a) The acceleration at \( t = 6.0 \, \text{s} \)
- b) The distance traveled in \( 8.0 \, \text{s} \)

### Part a: Acceleration at \( t = 6.0 \, \text{s} \)
- From the graph, at \( t = 6.0 \, \text{s} \), the particle's velocity is decreasing. We can see that this corresponds to a straight line, indicating constant negative acceleration.
- The slope of the velocity-time graph represents the acceleration.

To find the acceleration, we calculate the slope in the region between \( t = 4.0 \, \text{s} \) and \( t = 7.0 \, \text{s} \):
- Initial velocity at \( t = 4.0 \, \text{s} \) is \( 8 \, \text{m/s} \)
- Final velocity at \( t = 7.0 \, \text{s} \) is \( -8 \, \text{m/s} \)

The change in velocity \( \Delta v \) is:
\[
\Delta v = -8 \, \text{m/s} - 8 \, \text{m/s} = -16 \, \text{m/s}
\]

The change in time \( \Delta t \) is:
\[
\Delta t = 7.0 \, \text{s} - 4.0 \, \text{s} = 3.0 \, \text{s}
\]

Thus, the acceleration \( a \) is:
\[
a = \frac{\Delta v}{\Delta t} = \frac{-16 \, \text{m/s}}{3.0 \, \text{s}} = -5.33 \, \text{m/s}^2
\]

At \( t = 6.0 \, \text{s} \), the acceleration is \( -5.33 \, \text{m/s}^2 \).

### Part b: Distance Traveled in 8.0 s
- The area under the velocity-time graph gives the displacement.
- We will divide the graph into three regions: from 0 to 4 s, from 4 to 7 s, and from 7 to 8 s.

1. **From \( t = 0 \, \text{s} \) to \( t = 4 \, \text{s} \)**:
   - This is a triangle with base \( 4 \, \text{s} \) and height \( 8 \, \text{m/s} \).
   - Area (displacement) = \( \frac{1}{2} \times 4 \times 8 = 16 \, \text{m} \).

2. **From \( t = 4 \, \text{s} \) to \( t = 7 \, \text{s} \)**:
   - This is a triangle with base \( 3 \, \text{s} \) and height \( -8 \, \text{m/s} \).
   - Area (displacement) = \( \frac{1}{2} \times 3 \times (-8) = -12 \, \text{m} \).

3. **From \( t = 7 \, \text{s} \) to \( t = 8 \, \text{s} \)**:
   - This is a small triangle with base \( 1 \, \text{s} \) and height \( -8 \, \text{m/s} \).
   - Area (displacement) = \( \frac{1}{2} \times 1 \times (-8) = -4 \, \text{m} \).

The total distance traveled is the sum of the absolute values of the areas:
\[
\text{Total distance} = |16| + |12| + |4| = 16 + 12 + 4 = 32 \, \text{m}
\]

Thus, the distance traveled in \( 8.0 \, \text{s} \) is **32 meters**.

### Summary:
- a) The acceleration at \( t = 6.0 \, \text{s} \) is \( -5.33 \, \text{m/s}^2 \).
- b) The total distance traveled in \( 8.0 \, \text{s} \) is \( 32 \, \text{m} \).

---

Would you like more details or have any questions?

Here are some related questions for practice:
1. What would be the displacement in 8.0 seconds?
2. How would the acceleration change if the particle's final velocity was different?
3. Can you calculate the total change in velocity between \( t = 0 \, \text{s} \) and \( t = 8 \, \text{s} \)?
4. How would the total distance change if the acceleration in the second part was less?
5. What would happen if the particle continued moving past \( t = 8.0 \, \text{s} \)?

**Tip:** Always remember that the area under the velocity-time graph represents displacement, not total distance, unless you take absolute values of all areas.

A particle starts from rest and moves as shown in the velocity-time graph. Determine: a) Its acceleration at t = 6.0 s, and b) The distance covered in 8.0 s.

This problem involves finding the acceleration at t = 6.0 s and the total distance traveled in 8.0 s, based on a given velocity-time graph. The solution includes the use of kinematic formulas and graph interpretation techniques.

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