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The problem involves finding the relative velocity \( z \), which is the rate of separation between a car (mobil) and a train (kereta api) moving along perpendicular paths.

### Given dataThe problem involves finding the relative velocity \( z \), which is the rate of separation between a car (mobil) and a train (kereta api) moving along perpendicular paths.

### Given 
- The car is moving along the line labeled "mobil" with a velocity \( v_m = 30 \, \text{km/jam} \).
- The train is moving horizontally along the x-axis with a velocity \( v_k = 20 \, \text{km/jam} \).
- The positions of the car and train at time \( t = 0 \) are such that the distance between them is described by the coordinates of their paths.

### Coordinate System:
- \( x \) represents the position of the train along the x-axis.
- \( y \) represents the position of the car along the y-axis.
- \( z \) represents the distance between the car and the train (the hypotenuse of the right triangle formed by \( x \) and \( y \)).

### Objective:
- Find the rate of change of \( z \) (denoted as \( \frac{dz}{dt} \))—the rate of separation between the car and the train.
- Also, find \( x' \), \( y' \), and \( z' \), which represent the velocities along the respective axes.

### Step-by-Step Solution:

1. **Position Relations**:
   At any time \( t \), the positions of the train and car can be described as:
   \[
   x(t) = 1.5 + v_k \cdot t = 1.5 + 20 \cdot t
   \]
   \[
   y(t) = 0.5 + v_m \cdot t = 0.5 + 30 \cdot t
   \]
   
2. **Distance \( z(t) \)**:
   The distance \( z(t) \) is the hypotenuse of the right triangle formed by \( x(t) \) and \( y(t) \):
   \[
   z(t) = \sqrt{x(t)^2 + y(t)^2}
   \]

3. **Differentiating \( z(t) \)**:
   To find the rate of change of \( z \) with respect to time \( t \), apply the chain rule:
   \[
   \frac{dz}{dt} = \frac{1}{2} \cdot \left( \frac{2x(t) \cdot \frac{dx}{dt} + 2y(t) \cdot \frac{dy}{dt}}{z(t)} \right)
   \]
   Simplifying:
   \[
   \frac{dz}{dt} = \frac{x(t) \cdot v_k + y(t) \cdot v_m}{z(t)}
   \]
   where \( \frac{dx}{dt} = v_k = 20 \, \text{km/jam} \) and \( \frac{dy}{dt} = v_m = 30 \, \text{km/jam} \).

4. **Calculate at \( t = 5 \)**:
   Now, substitute \( t = 5 \) into the expressions for \( x(t) \), \( y(t) \), and \( z(t) \):
   \[
   x(5) = 1.5 + 20 \cdot 5 = 101.5 \, \text{km}
   \]
   \[
   y(5) = 0.5 + 30 \cdot 5 = 150.5 \, \text{km}
   \]
   \[
   z(5) = \sqrt{(101.5)^2 + (150.5)^2} = \sqrt{10302.25 + 22650.25} = \sqrt{32952.5} \approx 181.51 \, \text{km}
   \]

5. **Find \( \frac{dz}{dt} \)**:
   \[
   \frac{dz}{dt} = \frac{101.5 \cdot 20 + 150.5 \cdot 30}{181.51} = \frac{2030 + 4515}{181.51} = \frac{6545}{181.51} \approx 36.06 \, \text{km/jam}
   \]

### Final Results:
- The rate of separation between the car and the train \( z' \) is approximately \( 36.06 \, \text{km/jam} \).
- The velocities along the axes are \( x' = 20 \, \text{km/jam} \), \( y' = 30 \, \text{km/jam} \), and \( z' \approx 36.06 \, \text{km/jam} \).

Do you need further details or explanations?

### Follow-up questions:
1. What would happen if the velocities of the car and train were reversed?
2. How does the distance between the car and train change as time increases?
3. Can we express the problem in terms of relative motion in different frames of reference?
4. How would the result change if the initial positions were different?
5. What physical situations could this model be used to represent?

#### Tip: Always sketch a diagram for problems involving relative motion—it helps in visualizing and organizing the data.

Cari kecepatan berpisah antara mobil dan kereta api (z) dan hubungan x', y', dan z'.

Temukan bagaimana menghitung kecepatan berpisah antara mobil dan kereta api yang bergerak pada lintasan tegak lurus. Masalah ini menggunakan konsep kinematika dan turunan untuk mendapatkan nilai z dan hubungan antara x', y', dan z'. Cocok untuk siswa SMA.

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