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The problem you are working on deals with the displacement \( x \) of a particle moving along a straight line, which is given by the equation:

\[
x = t^2 - 4t
\]

where \( x \) is in meters and \( t \) is the time in seconds.

### Let's go over the questions:

1. **The displacement of the particle from O is zero at \( t = 0 \) and \( t = 4 \):**
   - This statement is true because we find when the displacement is zero by solving \( x = t^2 - 4t = 0 \). Factoring gives:
   \[
   t(t - 4) = 0
   \]
   So, \( t = 0 \) or \( t = 4 \), which means the particle is at the origin at \( t = 0 \) and \( t = 4 \) seconds.

2. **The displacement of the particle after 2 seconds:**
   - We substitute \( t = 2 \) into the equation \( x = t^2 - 4t \):
   \[
   x = 2^2 - 4(2) = 4 - 8 = -4 \, \text{meters}.
   \]
   This matches the provided answer of \(-4\) meters.

3. **The distance that the particle has travelled after 4 seconds:**
   - To find the total distance traveled, we need to consider the movement between turning points (where velocity is zero) and take absolute values.
   - First, the velocity is \( v = \frac{dx}{dt} = 2t - 4 \). Set \( v = 0 \) to find the turning point:
   \[
   2t - 4 = 0 \quad \Rightarrow \quad t = 2 \, \text{seconds}.
   \]
   - Now, calculate the displacement at \( t = 0 \), \( t = 2 \), and \( t = 4 \):
     - \( x(0) = 0^2 - 4(0) = 0 \),
     - \( x(2) = 2^2 - 4(2) = -4 \, \text{meters}, \)
     - \( x(4) = 4^2 - 4(4) = 0 \, \text{meters}. \)
   - The particle moves from \( 0 \) to \( -4 \) (a distance of 4 meters) and then back to \( 0 \) (another 4 meters).
   - Total distance traveled after 4 seconds is \( 4 + 4 = 8 \, \text{meters} \), not 0 meters as marked in your submission.

4. **The distance that the particle has travelled after 6 seconds:**
   - Calculate displacement at \( t = 6 \):
   \[
   x(6) = 6^2 - 4(6) = 36 - 24 = 12 \, \text{meters}.
   \]
   - Between \( t = 4 \) and \( t = 6 \), the displacement goes from 0 to 12 meters, so the distance covered in this interval is 12 meters.
   - Total distance traveled from \( t = 0 \) to \( t = 6 \) is the 8 meters already covered (from \( t = 0 \) to \( t = 4 \)) plus 12 meters, giving \( 8 + 12 = 20 \, \text{meters} \).

### Correction summary:
- **Question 3:** The total distance after 4 seconds should be **8 meters**.
- **Question 4:** The total distance after 6 seconds should be **20 meters**, not 12 meters.

---

Would you like a more detailed breakdown or have any questions? Here are some further questions to consider:

1. What is the velocity of the particle at \( t = 2 \) and \( t = 6 \)?
2. Can you calculate the acceleration of the particle at any given time?
3. How can you find the time when the particle reaches its maximum displacement?
4. What would happen if the equation for displacement was changed to \( x = t^2 - 5t \)?
5. How would you compute the total displacement, not just distance traveled, after 6 seconds?

**Tip:** To calculate the total distance traveled, always account for direction changes by calculating displacement between turning points and summing the absolute values.

The displacement from O of a particle moving along a straight line is given by x = t^2 - 4t, where x is measured in meters. Find: (1) The times when the displacement is zero. (2) The displacement of the particle after 2 seconds. (3) The distance the particle has travelled after 4 seconds. (4) The distance the particle has travelled after 6 seconds.

Solve a kinematics problem involving the displacement function x = t^2 - 4t. Learn how to find the times when displacement is zero, compute displacement at specific times, and calculate the total distance traveled by the particle over time intervals. This solution covers concepts like differentiation and absolute values.

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