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In the image, the problem involves calculating the speed of separation (denoted as \( z \)) between a car (mobil) and a train (kereta api), and determining the rates of change of the \( x' \), \( y' \), and \( z' \) components over time.

### Given:
- The car moves at a speed of \( v_m = 30 \) km/h.
- The train moves at a speed of \( v_k = 20 \) km/h.
- At \( t = 0 \), the car is positioned 0.5 km from the \( y \)-axis and the train is 1.5 km away along the \( x \)-axis.
- The car and the train are moving perpendicular to each other (the car moves along the \( y \)-axis and the train moves along the \( x \)-axis).

### Approach:
Let’s define the following:
- \( x(t) \) represents the position of the train along the \( x \)-axis, and \( y(t) \) represents the position of the car along the \( y \)-axis.
- The distance \( z(t) \) between the car and the train is the hypotenuse of a right triangle, with sides \( x(t) \) and \( y(t) \).

At any time \( t \), the distance between the car and the train can be found using the Pythagorean theorem:
\[
z(t) = \sqrt{x(t)^2 + y(t)^2}
\]

The rates \( x'(t) \), \( y'(t) \), and \( z'(t) \) are the time derivatives of the positions and the distance, respectively. To find them, we can use:

1. **Position Functions**:
   - \( x(t) = 1.5 - v_k t = 1.5 - 20t \) (since the train is moving towards the origin along the \( x \)-axis).
   - \( y(t) = 0.5 + v_m t = 0.5 + 30t \) (since the car is moving away from the origin along the \( y \)-axis).

2. **Velocity Components**:
   - \( x'(t) = -v_k = -20 \) km/h (negative because the train is moving towards the origin).
   - \( y'(t) = v_m = 30 \) km/h.

3. **Rate of Change of \( z(t) \)**:
   Using the chain rule, we can compute \( z'(t) \) as follows:
   \[
   z'(t) = \frac{d}{dt} \left( \sqrt{x(t)^2 + y(t)^2} \right) = \frac{x(t) x'(t) + y(t) y'(t)}{z(t)}
   \]
   This gives the rate at which the distance between the car and the train changes.

Let me know if you'd like me to compute these values for a specific time \( t \) or if you have any questions!

### Expanding on this:
1. How do you find the initial distance \( z(0) \) between the car and the train?
2. What would happen to \( z(t) \) if the train or car speeds were different?
3. Can you derive the time when \( z'(t) = 0 \), meaning the separation distance stops changing momentarily?
4. How does the relationship between \( x'(t) \), \( y'(t) \), and \( z'(t) \) depend on their directions of motion?
5. Can you calculate the angle of separation between the car and the train’s trajectory at any given time?

#### Tip:
In problems like this, breaking the motions into perpendicular components (like \( x \) and \( y \)) helps simplify the calculation of distances and rates of change!

Find the separation speed (z) between a car and a train, and calculate x', y', and z'. The car moves at 30 km/h and the train at 20 km/h. Initially, the car is 0.5 km from the y-axis, and the train is 1.5 km from the x-axis. The car and the train move perpendicularly to each other.

Learn how to calculate the separation speed between a car moving at 30 km/h and a train moving at 20 km/h using kinematics and the Pythagorean theorem. This problem involves finding the rate of change of distance using calculus principles, suitable for Grades 10-12 students.

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