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 steps:

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

### Step 3: Analyze the velocity intervals
- For \( t < 2 \, \text{s} \), the velocity is negative (the particle is moving in the negative direction).
- At \( t = 2 \, \text{s} \), the velocity is zero (the particle momentarily stops).
- For \( t > 2 \, \text{s} \), the velocity is positive (the particle moves in the positive direction).

### Step 4: Calculate displacement from \( t = 0 \) to \( t = 3.5 \, \text{s} \)
We need to break the calculation into two intervals: from \( t = 0 \) to \( t = 2 \) and from \( t = 2 \) to \( t = 3.5 \).

#### Interval 1: From \( t = 0 \) to \( t = 2 \)
The displacement \( s_1 \) is given by integrating the velocity:
\[ s_1 = \int_0^2 (3t^2 - 6t) \, dt \]
Calculating the integral:
\[ s_1 = \left[ t^3 - 3t^2 \right]_0^2 = (2^3 - 3 \cdot 2^2) - (0 - 0) = 8 - 12 = -4 \, \text{m} \]

#### Interval 2: From \( t = 2 \) to \( t = 3.5 \)
Now we calculate the displacement \( s_2 \):
\[ s_2 = \int_2^{3.5} (3t^2 - 6t) \, dt \]
Calculating the integral:
\[ s_2 = \left[ t^3 - 3t^2 \right]_2^{3.5} = \left( (3.5)^3 - 3(3.5)^2 \right) - \left( 2^3 - 3(2^2) \right) \]
Calculating each part:
\[ (3.5)^3 = 42.875, \quad 3(3.5)^2 = 36.75 \]
So,
\[ s_2 = (42.875 - 36.75) - (-4) = 6.125 + 4 = 10.125 \, \text{m} \]

### Step 5: Total displacement
The total displacement from \( t = 0 \) to \( t = 3.5 \) is:
\[ \text{Total displacement} = s_1 + s_2 = -4 + 10.125 = 6.125 \, \text{m} \]

### Step 6: Calculate total distance travelled
The total distance travelled is the sum of the absolute values of the displacements:
\[ \text{Total distance} = |s_1| + |s_2| = 4 + 10.125 = 14.125 \, \text{m} \]

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

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

### Final Answers
- Distance travelled: \( 14.125 \, \text{m} \)
- Average velocity: \( 1.75 \, \text{m/s} \)
- Average speed: \( 4.03 \, \text{m/s} \)