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To solve the problem, we need to analyze the given velocity-time graph and the corresponding options. The velocity-time graph shows how the velocity of a particle changes over time. The particle's displacement can be found by calculating the area under the curve of the velocity-time graph.

### Steps to Solve:

1. **Determine the position at \( t = 0 \):**
   The initial position of the particle is given as \( x = -16 \, \text{m} \) at \( t = 0 \).

2. **Find the areas under the velocity-time graph:**
   - **From \( t = 0 \) to \( t = 10 \) seconds:**
     The area is a triangle with a base of 10 seconds and a height of 4 m/s.
     \[
     \text{Area} = \frac{1}{2} \times 10 \times 4 = 20 \, \text{m}
     \]
   - **From \( t = 10 \) to \( t = 18 \) seconds:**
     The area is a rectangle with a base of 8 seconds and a height of 4 m/s.
     \[
     \text{Area} = 8 \times 4 = 32 \, \text{m}
     \]
   - **From \( t = 18 \) to \( t = 24 \) seconds:**
     The area is a triangle with a base of 6 seconds and a height of 4 m/s.
     \[
     \text{Area} = \frac{1}{2} \times 6 \times 4 = 12 \, \text{m}
     \]
   - **From \( t = 24 \) to \( t = 30 \) seconds:**
     The area is a triangle with a base of 6 seconds and a height of 4 m/s (but it's negative since the velocity is negative).
     \[
     \text{Area} = -\frac{1}{2} \times 6 \times 4 = -12 \, \text{m}
     \]
   - **From \( t = 30 \) to \( t = 40 \) seconds:**
     The area is a triangle with a base of 10 seconds and a height of 6 m/s (negative).
     \[
     \text{Area} = -\frac{1}{2} \times 10 \times 6 = -30 \, \text{m}
     \]

3. **Calculate the position of the particle at different times:**
   - At \( t = 10 \, \text{s} \):
     \[
     x_{10} = -16 + 20 = 4 \, \text{m}
     \]
   - At \( t = 18 \, \text{s} \):
     \[
     x_{18} = 4 + 32 = 36 \, \text{m}
     \]
   - At \( t = 24 \, \text{s} \):
     \[
     x_{24} = 36 + 12 = 48 \, \text{m}
     \]
   - At \( t = 30 \, \text{s} \):
     \[
     x_{30} = 48 - 12 = 36 \, \text{m}
     \]
   - At \( t = 40 \, \text{s} \):
     \[
     x_{40} = 36 - 30 = 6 \, \text{m}
     \]

4. **Analyze the options:**
   - **Option A: The maximum value of the position coordinate is 54 m.**
     - **Incorrect**. The maximum value is 48 m.
   - **Option B: The maximum value of the position coordinate is 36 m.**
     - **Incorrect**. The maximum value is 48 m.
   - **Option C: The particle is at the position of 36 m at \( t = 18 \, \text{s} \).**
     - **Correct**. As calculated, at \( t = 18 \, \text{s} \), the position is 36 m.
   - **Option D: The particle is at the position 36 m at \( t = 30 \, \text{s} \).**
     - **Correct**. As calculated, at \( t = 30 \, \text{s} \), the position is 36 m.

### Conclusion:
The correct answers are **C** and **D**.

---

Would you like further details or have any questions?

### Related Questions:
1. How would the result change if the initial position was different?
2. What would the position be at \( t = 20 \) seconds?
3. How would the velocity-time graph affect the maximum displacement?
4. What is the significance of negative velocity in the graph?
5. How can we determine the total distance traveled by the particle?

### Tip:
Always calculate the area under the velocity-time graph carefully, as it directly relates to the displacement of the particle over time.

Learn how to analyze a velocity-time graph to calculate displacement. This problem involves finding areas under the curve and determining the particle's position at different times. Suitable for high school students studying physics or mathematics.

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