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.

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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.