Solving the Train Length Problem

This problem involves a train moving at a constant speed. When a train crosses a point object like a pole, the distance it covers is equal to its own length. When it crosses a platform or a bridge, the distance it covers is equal to the sum of its own length and the length of the platform or bridge.

Understanding the Concepts:

• Crossing a pole: The train covers a distance equal to its length.
• Crossing a bridge: The train covers a distance equal to its length plus the length of the bridge.
• The speed of the train is constant in both scenarios.

Setting up the Equations

Let the length of the train be \(L\) meters and its speed be \(S\) meters per second.

We are given the following information:

• Time to cross a pole (\(T_{pole}\)) = 10 seconds
• Time to cross the bridge (\(T_{bridge}\)) = 25 seconds
• Length of the bridge (\(L_{bridge}\)) = 75 meters

Using the formula: Speed \(= \frac{\text{Distance}}{\text{Time}}\)

Case 1: Crossing the pole

The distance covered is the length of the train, \(L\).

Speed \(S = \frac{L}{T_{pole}}\)

\(S = \frac{L}{10}\) (Equation 1)

Case 2: Crossing the bridge

The distance covered is the length of the train plus the length of the bridge, \(L + L_{bridge}\).

Distance = \(L + 75\)

Speed \(S = \frac{\text{Distance}}{T_{bridge}}\)

\(S = \frac{L + 75}{25}\) (Equation 2)

Solving for the Length of the Train

Since the speed of the train is constant, we can equate the expressions for \(S\) from Equation 1 and Equation 2:

\(\frac{L}{10} = \frac{L + 75}{25}\)

Now, we solve this equation for \(L\):

• Multiply both sides by the least common multiple of 10 and 25, which is 50, or simply cross-multiply.
• \(25 \times L = 10 \times (L + 75)\)
• \(25L = 10L + 10 \times 75\)
• \(25L = 10L + 750\)
• Subtract \(10L\) from both sides:
• \(25L - 10L = 750\)
• \(15L = 750\)
  
• Divide both sides by 15:
• \(L = \frac{750}{15}\)
• \(L = 50\)

The length of the train is 50 meters.

Verification

If the length of the train is 50 m, its speed when crossing the pole in 10 seconds is:

\(S = \frac{50 \text{ m}}{10 \text{ s}} = 5 \text{ m/s}\)

When crossing the bridge, the total distance is \(50 \text{ m} + 75 \text{ m} = 125 \text{ m}\). The time taken with a speed of 5 m/s is:

\(T_{bridge} = \frac{125 \text{ m}}{5 \text{ m/s}} = 25 \text{ s}\)

This matches the given time for crossing the bridge, confirming our calculation is correct.

Thus, the length of the train is 50 meters.

Revision Table: Train Problems Basics

Additional Information: Speed, Time, and Distance

The relationship between speed, time, and distance is fundamental to solving train problems. The core formula is:

Speed \(= \frac{\text{Distance}}{\text{Time}}\)

This can be rearranged to find distance or time:

• Distance = Speed \(\times\) Time
• Time = \(\frac{\text{Distance}}{\text{Speed}}\)

Units must be consistent. If speed is in km/hr, distance should be in km and time in hours. If speed is in m/s, distance should be in meters and time in seconds.

To convert units:

• km/hr to m/s: Multiply by \(\frac{5}{18}\)
• m/s to km/hr: Multiply by \(\frac{18}{5}\)

Understanding these relationships and the specific distances covered when a train crosses different objects is key to solving various problems in this topic.