Understanding the Train Crossing Problem

This problem involves a train crossing a tunnel and a bridge. When a train crosses an object like a tunnel or a bridge, the total distance covered by the train is the length of the train plus the length of the object it is crossing. We are given the time taken for the train to cross a tunnel of a specific length and a bridge of another length. Using this information, we can determine the train's speed and its length.

Setting up Equations for Train Speed and Length

Let the length of the train be $L$ meters and the speed of the train be $V$ meters per second (m/s).

Scenario 1: Crossing the tunnel

The train crosses a tunnel of length 600 m in 54 seconds.

Distance covered = Length of train + Length of tunnel = $L + 600$ m

Time taken = 54 seconds

Using the formula: Speed = Distance / Time

We get Equation (1): $$V = \frac{L + 600}{54}$$

Scenario 2: Crossing the bridge

The train crosses a bridge of length 350 m in 36 seconds.

Distance covered = Length of train + Length of bridge = $L + 350$ m

Time taken = 36 seconds

Using the formula: Speed = Distance / Time

We get Equation (2): $$V = \frac{L + 350}{36}$$

Solving for Train Length and Speed

Since the speed of the train is the same in both scenarios, we can equate Equation (1) and Equation (2):

$$\frac{L + 600}{54} = \frac{L + 350}{36}$$

To solve for $L$, we can cross-multiply and simplify. Notice that 54 and 36 share a common factor of 18 ($54 = 18 \times 3$, $36 = 18 \times 2$). We can simplify the denominators first:

$$\frac{L + 600}{3} = \frac{L + 350}{2}$$

Now, cross-multiply:

$$2(L + 600) = 3(L + 350)$$

Distribute the numbers on both sides:

$$2L + 1200 = 3L + 1050$$

Rearrange the terms to isolate $L$:

$$1200 - 1050 = 3L - 2L$$

$$150 = L$$

So, the length of the train is 150 meters.

Now that we have the length of the train ($L = 150$ m), we can find the speed $V$ using either Equation (1) or Equation (2). Let's use Equation (2):

$$V = \frac{L + 350}{36}$$

Substitute $L = 150$:

$$V = \frac{150 + 350}{36}$$

$$V = \frac{500}{36}$$

Simplify the fraction:

$$V = \frac{250}{18} = \frac{125}{9} \text{ m/s}$$

The speed of the train is $125/9$ m/s.

Checking the Statements

Now let's check the given statements based on our calculations.

Statement (i): The speed of the train is 60 km/h.

Our calculated speed is $V = 125/9$ m/s. We need to convert this speed from m/s to km/h. To convert m/s to km/h, we multiply by $18/5$.

Speed in km/h $$= \left(\frac{125}{9}\right) \times \left(\frac{18}{5}\right)$$

$$= \left(\frac{125}{5}\right) \times \left(\frac{18}{9}\right)$$

$$= 25 \times 2$$

$$= 50 \text{ km/h}$$

The calculated speed is 50 km/h, not 60 km/h. Therefore, statement (i) is incorrect.

Statement (ii): The length of the train is 150 m.

Our calculated length of the train is $L = 150$ m. Therefore, statement (ii) is correct.

Conclusion on Statements

Based on our analysis:

• Statement (i) is incorrect. The train's speed is 50 km/h.
• Statement (ii) is correct. The train's length is 150 m.

Thus, only statement (ii) is correct.

Revision Table: Key Details

Additional Information: Speed, Distance, and Time Concepts

The relationship between speed, distance, and time is fundamental in solving such problems.

• Speed = Distance / Time
• Distance = Speed × Time
• Time = Distance / Speed

When a train crosses a point object (like a pole or a person), the distance covered is equal to the length of the train itself.

When a train crosses a linear object of some length (like a bridge, tunnel, or platform), the distance covered is the sum of the length of the train and the length of the object.

It is crucial to maintain consistent units for speed, distance, and time throughout the calculation. The standard units are meters (m) for distance, seconds (s) for time, and meters per second (m/s) for speed. If speed is given in kilometers per hour (km/h), it often needs to be converted to m/s for calculations involving meters and seconds. The conversion factors are:

• 1 km/h = $5/18$ m/s
• 1 m/s = $18/5$ km/h

In this problem, we used time in seconds and lengths in meters, calculated the speed in m/s, and then converted it to km/h to check statement (i).