The equation for the motion of the first car with respect to the Earth's surface is:

x₁ = v₁t + x₀

Where x₁ is the position of the car at time t, v₁ is the velocity of the car with respect to the Earth's surface, and x₀ is the initial position of the car.

The equation for the motion of the second car with respect to the first car is:

x₂ = v₂t + x₀

Where x₂ is the position of the second car at time t with respect to the first car, and v₂ is the velocity of the second car with respect to the first car.

To find the motion of the second car with respect to the Earth, we need to use the relative velocity formula:

v_{2, E} = v₂ + v_{1, E}

Where v_{2, E} is the velocity of the second car with respect to the Earth, and v_{1, E} is the velocity of the first car with respect to the Earth.

Since the velocity of the first car with respect to the Earth is given as 2 m/s along the x-axis, and the velocity of the second car with respect to the first car is not specified, we can use any value for v₂. Let's choose a value of 3 m/s along the y-axis, as this will result in a more interesting graph.

Thus, the velocity of the second car with respect to the Earth is:

v_{2, E} = (0, 3) m/s

This means that the velocity of the second car with respect to the Earth has a magnitude of 3 m/s and a direction of 90°, or directly above the first car's position.

Now, we can use this velocity in the equation for the motion of the second car with respect to the Earth:

x_{2, E} = v_{2, E}t + x₀

Since the initial position of the second car is the same as the initial position of the first car, we can set x₀ as 0.

Therefore, the equation for the motion of the second car with respect to the Earth is:

x_{2, E} = 3t

This is the graph of the second car's motion in the Earth's reference frame, with the origin at the initial position of the first car.

To graph the first car's motion in the Earth's reference frame, we can simply use its original velocity, which is 2 m/s along the x-axis. The equation for its motion is:

x_{1, E} = 2t

This results in a graph with a slope of 2, or a straight line.

The thrust force of each of the four engines can be calculated using the formula:

F = m*a

Where F is the thrust force, m is the mass of the airplane, and a is the acceleration.

Given that the airplane has a mass of 75 t and the resultant force is 120 kN, we can calculate the acceleration:

a = F/m = 120000 N / 75000 kg = 1.6 m/s^2

The frictional force is given by the formula:

F_f = u*m*a

Where F_f is the frictional force, u is the coefficient of friction, m is the mass of the airplane, and a is the acceleration.

Substituting the given values, we get:

F_f = 0.027 * 75000 kg * 1.6 m/s^2 = 3240 N

Since the resultant force of 120 kN is equal to the sum of the thrust and frictional forces, we can calculate the thrust force of each engine by subtracting the frictional force from the resultant force and dividing by 4 (since there are 4 engines).

Therefore, the thrust force of each engine is:

F_thrust = (120000 N - 3240 N)/4 = 29265 N

Congratulations, you now know the thrust force of each engine of the airplane during takeoff!

The maximum height that the 2 kg rock can reach can be determined using the equation h = (v²) / (2g), where h is the maximum height, v is the initial vertical velocity, and g is the acceleration due to gravity (9.81 m/s² on Earth). In this case, h = (10 m/s)² / (2*9.81 m/s²) = 1.02 meters.

Therefore, the rock will reach a maximum height of approximately 1.02 meters.

The acceleration of the train is 0.1 m/s^2, which can be calculated using the equation F=ma with the given values of force and mass. This means that the train is moving at a constant velocity since the acceleration is not changing.

To determine the tension in the couplings between the wagons, we need to analyze the forces acting on the train. Since the train is moving at a constant velocity, the force of friction must be equal and opposite to the force of the locomotive. Using the equation F=μmg, we can calculate this force to be 99.8 N (rounded to the nearest hundredth).

Next, we need to find the tension in the couplings between the wagons. Using the equation F=ma, we can calculate this to be 598 N (rounded to the nearest whole number). This force is acting in the opposite direction of the force of friction, which means that the couplings are experiencing a tension force.