This problem involves two trains moving in the same direction on parallel tracks. To solve this, we need to use the concept of relative speed and the relationship between distance, speed, and time.

Understanding Relative Speed for Trains in Same Direction

When two objects, like trains, move in the same direction, their relative speed is the difference between their individual speeds. This is because the faster train is effectively closing the distance to the slower train at a speed equal to the difference in their speeds.

Relative Speed $(\text{S}_{rel}) =$ Speed of Faster Train $-$ Speed of Slower Train

Calculating Relative Speed and Converting Units

The speeds are given in km/h, and the time is in minutes, while the required length is in meters. It's best to convert all units to meters and seconds (m/s) for consistency.

• Speed of Train 1 $(\text{S}_1) = 80 \text{ km/h}$
• Speed of Train 2 $(\text{S}_2) = 90 \text{ km/h}$

Since $\text{S}_2 > \text{S}_1$, the relative speed will be $\text{S}_2 - \text{S}_1$.

First, convert the speeds from km/h to m/s. The conversion factor is $\frac{5}{18}$ because $1 \text{ km} = 1000 \text{ m}$ and $1 \text{ hour} = 3600 \text{ seconds}$, so $1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} = \frac{10}{36} \text{ m/s} = \frac{5}{18} \text{ m/s}$.

• $\text{S}_1 = 80 \times \frac{5}{18} = \frac{400}{18} = \frac{200}{9} \text{ m/s}$
• $\text{S}_2 = 90 \times \frac{5}{18} = 5 \times 5 = 25 \text{ m/s}$

Now, calculate the relative speed:

$\text{S}_{rel} = \text{S}_2 - \text{S}_1 = 25 - \frac{200}{9}$

To subtract, find a common denominator:

$\text{S}_{rel} = \frac{25 \times 9}{9} - \frac{200}{9} = \frac{225}{9} - \frac{200}{9} = \frac{225 - 200}{9} = \frac{25}{9} \text{ m/s}$

Calculating Distance Covered During Crossing

When two trains cross each other (pass completely), the total distance covered relative to each other is equal to the sum of their lengths.

• Length of Train 1 $(\text{L}_1) = 230 \text{ m}$ (given)
• Length of Train 2 $(\text{L}_2) = ?$ (to be found)

Total distance $(\text{D}) = \text{L}_1 + \text{L}_2 = 230 + \text{L}_2$

Using Speed, Time, and Distance Formula

The trains crossed each other in 3 minutes. Convert this time to seconds.

• Time $(\text{T}) = 3 \text{ minutes} = 3 \times 60 \text{ seconds} = 180 \text{ seconds}$

The relationship between distance, speed, and time is:

$\text{Distance} = \text{Speed} \times \text{Time}$

In this case, the distance is the total length of the trains, the speed is the relative speed, and the time is the crossing time.

$\text{D} = \text{S}_{rel} \times \text{T}$

Substitute the values we have:

$230 + \text{L}_2 = \frac{25}{9} \times 180$

Simplify the right side of the equation:

$180 \div 9 = 20$

So, the equation becomes:

$230 + \text{L}_2 = 25 \times 20$

$230 + \text{L}_2 = 500$

Finding the Length of the Other Train

Now, solve for $\text{L}_2$:

$\text{L}_2 = 500 - 230$

$\text{L}_2 = 270$

The length of the other train is 270 meters.

The length of the other train is 270 m.

Revision Table: Train Relative Speed

Additional Information: Relative Speed Scenarios

Understanding relative speed is key in time and distance problems involving multiple moving objects. Here are some related scenarios:

• Trains Moving in Opposite Directions: If two trains move towards each other or away from each other, their relative speed is the sum of their individual speeds ($\text{S}_1 + \text{S}_2$).
• Train Crossing a Pole or a Person: When a train crosses a stationary object with negligible length (like a pole or a standing person), the distance covered by the train is equal to its own length. The speed is the train's speed.
• Train Crossing a Platform or Bridge: When a train crosses an object with significant length (like a platform, bridge, or tunnel), the distance covered by the train is equal to the sum of the train's length and the object's length. The speed is the train's speed.

These concepts are variations of the distance = speed × time formula, applied with the appropriate relative speed and total distance involved in the crossing.