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Let's break down the problem step-by-step.

### Given:
- Initial speed \( u = 0 \) km/hr (since the train is starting from rest)
- Final speed \( v = 250 \) km/hr
- Acceleration \( a = 3888 \) km/hr²
- Time \( h \) hours

The formula for speed under constant acceleration is:
\[
v = u + a \times t
\]
Since \( u = 0 \):
\[
v = a \times h
\]

Substitute the given values:
\[
250 = 3888 \times h
\]

To find the equation:
\[
250 = 3888h
\]

This equation matches option **B**.

So, the correct answer is:
- **B) 250 = 3,888h**

Would you like more details or have any questions?

### 5 Relative Questions:
1. How would the equation change if the train started at a non-zero initial speed?
2. How do you convert acceleration from km/hr² to m/s²?
3. What other kinematic equations can be used to solve problems involving constant acceleration?
4. How long would it take the train to reach a speed of 500 km/hr with the same acceleration?
5. What is the importance of understanding constant acceleration in real-life scenarios like train travel?

### Tip:
Always double-check the units given in a problem to ensure consistency when applying kinematic equations.

When an Alvia high-speed train that is stopped in Barcelona leaves the station, its speed increases (accelerates) at a constant rate of 3,888 kilometers per hour squared (km/hr^2). If h is the time, in hours, it takes for the train to reach a speed of 250 km/hr, which of the following equations best describes this situation?

Solve a kinematic problem involving a train accelerating at 3,888 km/hr^2 to reach a speed of 250 km/hr. This example covers key concepts in kinematics and algebra, ideal for students in Grades 8-10.

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