A passenger train and a goods train are running in the same direction on parallel railway tracks. If the passenger train now takes three times as long to pass the goods train, as when they are running in the opposite directions, then what is the ratio of the speed of the passenger train to that of the goods train?
2 : 1
Understanding the Train Problem: Relative Speeds
This problem involves two trains, a passenger train and a goods train, moving on parallel tracks. We are asked to find the ratio of their speeds based on the time it takes for one to pass the other in two different scenarios: running in the same direction and running in opposite directions.
The key concept here is relative speed. When two objects are moving, their relative speed is the speed at which the distance between them changes. The time it takes for one train to completely pass another is determined by the total distance (sum of their lengths) and their relative speed.
Defining Variables
- Let \(S_P\) be the speed of the passenger train.
- Let \(S_G\) be the speed of the goods train.
- Let \(L_P\) be the length of the passenger train.
- Let \(L_G\) be the length of the goods train.
The total distance that needs to be covered for one train to pass the other is the sum of their lengths, which is \(D = L_P + L_G\). This distance is constant for both scenarios (same direction and opposite directions).
Calculating Relative Speed in Different Scenarios
When the trains are running in the same direction, and the passenger train is passing the goods train, the passenger train must be faster than the goods train (\(S_P > S_G\)). The relative speed at which the passenger train gains on the goods train is the difference between their speeds:
Relative Speed (Same Direction) \(= S_P - S_G\)
When the trains are running in opposite directions, they are moving towards each other at a combined speed. The relative speed at which the distance between them decreases is the sum of their speeds:
Relative Speed (Opposite Directions) \(= S_P + S_G\)
Calculating Time Taken to Pass
The time taken for one train to pass the other is given by the total distance to be covered divided by the relative speed.
Time (Same Direction), \(T_{same} = \frac{\text{Total Distance}}{\text{Relative Speed (Same Direction)}} = \frac{L_P + L_G}{S_P - S_G}\)
Time (Opposite Directions), \(T_{opp} = \frac{\text{Total Distance}}{\text{Relative Speed (Opposite Directions)}} = \frac{L_P + L_G}{S_P + S_G}\)
Using the Given Information to Find the Ratio of Speeds
The problem states that the passenger train takes three times as long to pass the goods train when running in the same direction as when they are running in opposite directions. Mathematically, this can be written as:
\(T_{same} = 3 \times T_{opp}\)
Substitute the expressions for \(T_{same}\) and \(T_{opp}\):
\(\frac{L_P + L_G}{S_P - S_G} = 3 \times \frac{L_P + L_G}{S_P + S_G}\)
Since the term \((L_P + L_G)\) is common and non-zero on both sides of the equation, we can cancel it out:
\(\frac{1}{S_P - S_G} = \frac{3}{S_P + S_G}\)
Now, cross-multiply to solve for the relationship between \(S_P\) and \(S_G\):
\(1 \times (S_P + S_G) = 3 \times (S_P - S_G)\)
\(S_P + S_G = 3S_P - 3S_G\)
Rearrange the terms to group \(S_P\) terms and \(S_G\) terms:
\(S_G + 3S_G = 3S_P - S_P\)
\(4S_G = 2S_P\)
To find the ratio of the speed of the passenger train to that of the goods train, we want to find \(S_P / S_G\). Divide both sides by \(2S_G\):
\(\frac{4S_G}{2S_G} = \frac{2S_P}{2S_G}\)
\(\frac{4}{2} = \frac{S_P}{S_G}\)
\(2 = \frac{S_P}{S_G}\)
So, the ratio of the speed of the passenger train to that of the goods train, \(S_P : S_G\), is \(2 : 1\).
| Scenario | Relative Speed | Time Taken |
|---|---|---|
| Same Direction | \(S_P - S_G\) | \(T_{same} = \frac{L_P + L_G}{S_P - S_G}\) |
| Opposite Directions | \(S_P + S_G\) | \(T_{opp} = \frac{L_P + L_G}{S_P + S_G}\) |
Given \(T_{same} = 3 \times T_{opp}\)
\(\frac{L_P + L_G}{S_P - S_G} = 3 \times \frac{L_P + L_G}{S_P + S_G}\)
\(\frac{1}{S_P - S_G} = \frac{3}{S_P + S_G}\)
\(S_P + S_G = 3(S_P - S_G)\)
\(S_P + S_G = 3S_P - 3S_G\)
\(4S_G = 2S_P\)
\(\frac{S_P}{S_G} = \frac{4}{2} = \frac{2}{1}\)
The ratio is 2:1.
Revision Table: Key Concepts
| Concept | Description | Formula Used |
|---|---|---|
| Relative Speed (Same Direction) | Speed difference when objects move in the same direction. Used when a faster object overtakes a slower one. | \(S_1 - S_2\) (assuming \(S_1 > S_2\)) |
| Relative Speed (Opposite Directions) | Sum of speeds when objects move towards or away from each other. | \(S_1 + S_2\) |
| Time to Pass/Cross | Total distance (sum of lengths) divided by relative speed. | \(\frac{\text{Distance}}{\text{Relative Speed}}\) |
Additional Information: Train Problems
Train problems are a common type of question involving time, speed, and distance. They often involve relative speed and the concept that for a train to pass a point, pole, or another train, it must cover a distance equal to its own length or the sum of its length and the length of the other object.
- When a train passes a point object (like a pole, a person, or a signal), the distance covered is the length of the train itself.
- When a train passes a platform, bridge, or tunnel, the distance covered is the length of the train plus the length of the platform/bridge/tunnel.
- When a train passes another train, the distance covered is the sum of the lengths of the two trains.
- Relative speed is crucial when two trains are moving.
Understanding relative speed is key to solving these problems efficiently. Remember that speed, distance, and time are related by the formula: Distance = Speed × Time.
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