A particle moves along a horizontal path, such that its velocity is given by `v = (3t^2 - 6t) m//s,` where t is the time in seconds. If it is initially located at the origin O, determine the distance travelled by the particle in time interval from `t = 0` to `t = 3.5 s` and the particle's average velocity and average speed during the same time interval.

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Determine the velocity function
The velocity of the particle is given by:
\[ v(t) = 3t^2 - 6t \, \text{m/s} \]

### Step 2: Find the time intervals when the velocity is zero
To find when the particle changes direction, we need to set the velocity function to zero:
\[ 3t^2 - 6t = 0 \]
Factoring out \(3t\):
\[ 3t(t - 2) = 0 \]
This gives us:
\[ t = 0 \quad \text{or} \quad t = 2 \]

### Step 3: Analyze the velocity function
- For \( t < 0 \): Not applicable since time cannot be negative.
- For \( 0 < t < 2 \): The velocity is negative (the particle moves in the negative direction).
- For \( t > 2 \): The velocity is positive (the particle moves in the positive direction).

### Step 4: Calculate the displacement from \( t = 0 \) to \( t = 3.5 \)
We will calculate the displacement in two parts:
1. From \( t = 0 \) to \( t = 2 \) (where the velocity is negative).
2. From \( t = 2 \) to \( t = 3.5 \) (where the velocity is positive).

#### Part 1: Displacement from \( t = 0 \) to \( t = 2 \)
The displacement \( x \) can be calculated by integrating the velocity function:
\[ x = \int_0^2 (3t^2 - 6t) \, dt \]
Calculating the integral:
\[ x = \left[ t^3 - 3t^2 \right]_0^2 = (2^3 - 3 \cdot 2^2) - (0 - 0) = (8 - 12) = -4 \, \text{m} \]
This means the particle moves 4 meters in the negative direction.

#### Part 2: Displacement from \( t = 2 \) to \( t = 3.5 \)
Now, we calculate the displacement from \( t = 2 \) to \( t = 3.5 \):
\[ x = \int_2^{3.5} (3t^2 - 6t) \, dt \]
Calculating the integral:
\[ x = \left[ t^3 - 3t^2 \right]_2^{3.5} = \left( (3.5)^3 - 3(3.5)^2 \right) - \left( 2^3 - 3(2^2) \right) \]
Calculating each part:
\[ = \left( 42.875 - 36.75 \right) - (8 - 12) = 6.125 + 4 = 10.125 \, \text{m} \]

### Step 5: Total displacement
The total displacement from \( t = 0 \) to \( t = 3.5 \):
\[ \text{Total Displacement} = -4 + 10.125 = 6.125 \, \text{m} \]

### Step 6: Calculate the total distance traveled
The total distance traveled is the sum of the absolute values of the displacements:
\[ \text{Total Distance} = | -4 | + | 10.125 | = 4 + 10.125 = 14.125 \, \text{m} \]

### Step 7: Calculate the average velocity
The average velocity is given by:
\[ \text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{6.125 \, \text{m}}{3.5 \, \text{s}} = 1.75 \, \text{m/s} \]

### Step 8: Calculate the average speed
The average speed is given by:
\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{14.125 \, \text{m}}{3.5 \, \text{s}} \approx 4.03 \, \text{m/s} \]

### Final Results
- **Total Distance Traveled**: \( 14.125 \, \text{m} \)
- **Average Velocity**: \( 1.75 \, \text{m/s} \)
- **Average Speed**: \( 4.03 \, \text{m/s} \)