Step 1: Understanding the Concept:
When solving problems related to a train crossing an object, the "relative distance" is key.
If a train crosses a point object (like a pole or a standing person), the distance covered is just the train's length.
However, when a train crosses an object with its own significant length (like a platform, bridge, or tunnel), the train is not considered to have "crossed" the object until its very last carriage has passed the far end of that object.
Therefore, the total distance the engine travels from the moment it enters the platform until the last car leaves it is the sum of the train's length and the platform's length.
Also, since dimensions are in meters and time in seconds, we find speed in \(m/s\). Most exam options require speed in \(km/h\), so unit conversion is mandatory.
Step 2: Key Formula or Approach:
1. \(\text{Total Distance} = \text{Length of Train} + \text{Length of Platform}\)
2. \(\text{Speed (in m/s)} = \frac{\text{Total Distance}}{\text{Total Time Taken}}\)
3. To convert speed from \(m/s\) to \(km/h\), multiply the value by \(\frac{18}{5}\).
Step 3: Detailed Explanation:
Let's list the given parameters:
Length of the train = \(240\) m
Length of the platform = \(360\) m
Time taken (\(t\)) = \(30\) seconds
First, we calculate the total distance that the train needs to cover to fully clear the platform.
\[ \text{Total Distance (D)} = 240 + 360 = 600 \text{ meters} \]
Now, we calculate the speed of the train in the standard metric unit of meters per second.
\[ \text{Speed (v)} = \frac{\text{Distance}}{\text{Time}} \]
\[ v = \frac{600}{30} = 20 \text{ m/s} \]
The options provided in the question are all in kilometers per hour (\(km/h\)). We must convert \(20 \text{ m/s}\) using the conversion factor \(\frac{18}{5}\).
(Note: This factor comes from \(\frac{1 \text{ km}}{1 \text{ hour}} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}\)).
\[ \text{Speed in km/h} = 20 \times \left( \frac{18}{5} \right) \]
Simplifying the expression:
\[ \text{Speed} = \left( \frac{20}{5} \right) \times 18 \]
\[ \text{Speed} = 4 \times 18 = 72 \text{ km/h} \]
Thus, the train is traveling at a constant speed of \(72 \text{ km/h}\).
Step 4: Final Answer:
The speed of the train is 72 km/h.