To determine the maximum speed of the rocket, we need to calculate the change in velocity due to the thrust force applied over the 30 seconds. Here's the step-by-step process:

Step 1: Relation Between Force and Momentum

The thrust force $F_x(t)$ generates an acceleration, which changes the velocity of the rocket. Using Newton's second law:

$F_x = m \cdot a \quad \Rightarrow \quad a = \frac{F_x}{m}$

Since acceleration is the derivative of velocity, the velocity change can be found by integrating the force over time:

$\Delta v = \frac{1}{m} \int_0^{30} F_x(t) \, dt$

Step 2: Area Under the Force-Time Graph

The thrust force $F_x(t)$ is represented as a triangular function, so its integral corresponds to the area under the graph. The triangle has a base of 30 s and a height of 1000 N. Its area is:

$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 30 \cdot 1000 = 15000 \, \text{N·s}$

This area represents the total impulse delivered by the thrust force.

Step 3: Calculate the Velocity Change

Impulse is related to the change in momentum:

$\text{Impulse} = \Delta p = m \cdot \Delta v$

Thus, the change in velocity is:

$\Delta v = \frac{\text{Impulse}}{m} = \frac{15000}{425} \approx 35.29 \, \text{m/s}$

Step 4: Calculate the Maximum Speed

The initial speed of the rocket is $v_0 = 85 \, \text{m/s}$. Adding the change in velocity:

$v_{\text{max}} = v_0 + \Delta v = 85 + 35.29 = 120.29 \, \text{m/s}$

Final Answer:

The maximum speed of the rocket is approximately:

$\boxed{120.3 \, \text{m/s}}$

Do you want details about a specific part of this calculation or any additional questions? Here are some related questions:

1. How does impulse relate to the force-time graph?
2. What would happen to the velocity if the mass of the rocket were reduced?
3. How is the integration of a linear force graph performed explicitly?
4. How would the maximum speed change if the thrust duration were shorter?
5. What role does the initial velocity play in determining the maximum speed?

Tip: For force-time graphs, the area under the curve always corresponds to the impulse, which simplifies many calculations in mechanics.