Subway Train Journey Analysis

To determine the total distance between the two stations, we need to analyze the subway train's motion in three distinct phases:

1. The acceleration phase, where the train starts from rest and speeds up.
2. The constant speed phase, where the train moves at a steady velocity.
3. The deceleration phase, where the train slows down until it comes to a complete stop.

Subway Train Acceleration Phase

In the first phase of its journey, the subway train starts from rest and accelerates for a specific period. We need to calculate both the final velocity reached and the distance covered during this acceleration.

• Initial velocity (\(u\)) = \(0\) m/s (since the train starts from rest)
• Acceleration (\(a\)) = \(3\) m/s\(^2\)
• Time (\(t\)) = \(10\) s

First, we calculate the final velocity (\(v\)) of the subway train at the end of the acceleration phase using the first equation of motion:

\(v = u + at\)

\(v = 0 \text{ m/s} + (3 \text{ m/s}^2 \times 10 \text{ s})\)

\(v = 30\) m/s

Next, we calculate the distance covered (\(s_1\)) during this acceleration phase using the second equation of motion:

\(s_1 = ut + \frac{1}{2}at^2\)

\(s_1 = (0 \text{ m/s} \times 10 \text{ s}) + \frac{1}{2}(3 \text{ m/s}^2)(10 \text{ s})^2\)

\(s_1 = 0 + \frac{1}{2}(3)(100)\)

\(s_1 = \frac{300}{2}\)

\(s_1 = 150\) m

Subway Train Constant Speed Phase

After accelerating, the subway train maintains a constant speed for a period. This constant speed is the final velocity achieved at the end of the acceleration phase. Let's calculate the distance covered in this phase.

• Constant speed (\(v\)) = \(30\) m/s (from the end of the acceleration phase)
• Time (\(t\)) = \(20\) s

The distance covered (\(s_2\)) at constant speed is calculated by simply multiplying the speed by the time:

\(s_2 = \text{speed} \times \text{time}\)

\(s_2 = 30 \text{ m/s} \times 20 \text{ s}\)

\(s_2 = 600\) m

Subway Train Deceleration Phase

In the final phase, the subway train begins to decelerate until it comes to a complete stop at the next station. We need to find the distance covered during this deceleration.

• Initial velocity (\(u\)) = \(30\) m/s (the constant speed from the previous phase)
• Final velocity (\(v\)) = \(0\) m/s (since the train stops)
• Deceleration (\(a\)) = \(-5\) m/s\(^2\) (the negative sign indicates deceleration)

We use the third equation of motion, which relates initial velocity, final velocity, acceleration, and distance:

\(v^2 = u^2 + 2as\)

\(0^2 = (30 \text{ m/s})^2 + 2(-5 \text{ m/s}^2)s_3\)

\(0 = 900 - 10s_3\)

Rearranging the equation to solve for \(s_3\):

\(10s_3 = 900\)

\(s_3 = \frac{900}{10}\)

\(s_3 = 90\) m

Total Distance Between Stations

The total distance between the two stations is the sum of the distances covered in each of the three phases of the subway train's journey:

Total Distance = Distance during acceleration (\(s_1\)) + Distance at constant speed (\(s_2\)) + Distance during deceleration (\(s_3\))

Total Distance = \(150 \text{ m} + 600 \text{ m} + 90 \text{ m}\)

Total Distance = \(840\) m

Therefore, the distance between the two stations is 840 m.