Calculating Train Speed Passing Platform and Man

This problem involves a train passing two different objects: a platform and a man standing on the platform. The time taken for each event gives us information about the train's speed and length.

Understanding the Concepts

When a train passes a stationary object, the distance covered by the train depends on the object's length and the train's own length. The key idea is the point of reference.

• Passing a man or a point: When a train passes a man (who can be considered a point object with negligible length), the distance covered by the train is equal to the length of the train itself. This is because the time taken is from the moment the front of the train reaches the man until the moment the back of the train leaves the man.
• Passing a platform or a bridge: When a train passes a platform (which has a significant length), the distance covered by the train is equal to the length of the train plus the length of the platform. This is because the time taken is from the moment the front of the train reaches the start of the platform until the moment the back of the train leaves the end of the platform.

Setting up the Equations for Train Speed Calculation

Let:

• \(L\) be the length of the train in metres.
• \(S\) be the speed of the train in metres per second (m/s).

We are given:

• Length of the platform = 360 metres.
• Time taken to pass the platform = 40 seconds.
• Time taken to pass the man = 16 seconds.

We can use the formula: Distance = Speed × Time.

Scenario 1: Train passing the man

The distance covered is the length of the train, \(L\). The speed is \(S\) and the time is 16 seconds.

So, we have the equation:

\(L = S \times 16\)

\(L = 16S\) (Equation 1)

Scenario 2: Train passing the platform

The distance covered is the length of the train plus the length of the platform, \(L + 360\). The speed is \(S\) and the time is 40 seconds.

So, we have the equation:

\(L + 360 = S \times 40\)

\(L + 360 = 40S\) (Equation 2)

Solving for Train Speed

Now we have a system of two linear equations with two variables (\(L\) and \(S\)):

1. \(L = 16S\)
2. \(L + 360 = 40S\)

We can substitute the expression for \(L\) from Equation 1 into Equation 2:

\((16S) + 360 = 40S\)

Now, we solve for \(S\):

\(360 = 40S - 16S\)

\(360 = 24S\)

\(S = \frac{360}{24}\)

To simplify the fraction:

\(S = \frac{180}{12}\)

\(S = \frac{90}{6}\)

\(S = 15\)

So, the speed of the train is \(S = 15\) m/s.

Converting Speed from m/s to km/h

The options for the speed are given in kilometres per hour (kmph). We need to convert our speed from m/s to km/h.

The conversion factor is: \(1 \text{ m/s} = \frac{18}{5} \text{ km/h}\).

So, the speed in km/h is:

\(S_{\text{km/h}} = 15 \times \frac{18}{5}\)

\(S_{\text{km/h}} = (15 \div 5) \times 18\)

\(S_{\text{km/h}} = 3 \times 18\)

\(S_{\text{km/h}} = 54\)

Therefore, the speed of the train is 54 kmph.

This matches one of the given options.

Revision Table: Train Speed and Distance

Key formulas and conversions related to train problems:

Additional Information: Relative Speed Concepts

While this problem involved a stationary man and platform, train problems often involve relative speed when dealing with moving objects.

• Train passing a moving man/object (same direction): If the train and the man/object are moving in the same direction, the relative speed is the difference between their speeds (\(S_{\text{train}} - S_{\text{object}}\)). The distance covered is still the length of the train (if the object is a point) or length of train + length of object (if the object has length).
• Train passing a moving man/object (opposite direction): If the train and the man/object are moving in opposite directions, the relative speed is the sum of their speeds (\(S_{\text{train}} + S_{\text{object}}\)). The distance covered is again the length of the train (if the object is a point) or length of train + length of object (if the object has length).

Understanding relative speed is crucial for solving more complex train problems.