Step 1: Understanding the Question:
The problem requires us to find the individual running speeds of two distinct trains.
We are given the lengths of the two trains, which are 130 meters and 110 meters.
The problem outlines two specific scenarios regarding their motion.
When moving in the same direction, the faster train passes the slower one completely in one minute.
When moving in opposite directions, they cross each other completely in exactly 3 seconds.
Step 2: Key Formula or Approach:
The total distance to be covered for one train to completely pass another is always the sum of their individual lengths.
When two objects move in the same direction, their relative speed is the difference between their individual speeds ($V_1 - V_2$).
When two objects move in opposite directions, their relative speed is the sum of their individual speeds ($V_1 + V_2$).
We will formulate two simultaneous linear equations using the standard formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$.
Step 3: Detailed Explanation:
Let the speed of the faster train be $V_1$ m/s, and the speed of the slower train be $V_2$ m/s.
The total distance required to complete a pass in both scenarios is the combined length of the two trains.
Total distance $D = 130 \text{ m} + 110 \text{ m} = 240$ meters.
First, let us examine the scenario where they move in the same direction.
The time taken to pass is given as one minute, which we must convert to seconds: $1 \text{ minute} = 60 \text{ seconds}$.
The relative speed in this scenario is $V_1 - V_2$.
Using the speed formula, we write our first equation: \[ V_1 - V_2 = \frac{240}{60} \]
This simplifies neatly to: \[ V_1 - V_2 = 4 \text{ m/s} \quad \text{--- (Equation 1)} \]
Second, let us examine the scenario where they move in opposite directions.
The time taken to cross each other is given as exactly 3 seconds.
The relative speed in this scenario is $V_1 + V_2$.
Using the speed formula, we write our second equation: \[ V_1 + V_2 = \frac{240}{3} \]
This simplifies to: \[ V_1 + V_2 = 80 \text{ m/s} \quad \text{--- (Equation 2)} \]
Now, we have a simple system of linear equations to solve.
We can add Equation 1 and Equation 2 together to eliminate $V_2$.
\[ (V_1 - V_2) + (V_1 + V_2) = 4 + 80 \]
\[ 2V_1 = 84 \]
Dividing both sides by 2 gives the speed of the faster train.
\[ V_1 = 42 \text{ m/s} \]
To find the speed of the slower train, we substitute $V_1$ back into Equation 2.
\[ 42 + V_2 = 80 \]
\[ V_2 = 80 - 42 = 38 \text{ m/s} \]
The speeds of the trains are 42 m/s and 38 m/s respectively.
Step 4: Final Answer:
The speed of each train is 42 m/s and 38 m/s.