Train problems are totally based on four topics including conversion, distance formula, relativity, and train theory.

**Distance formula**: **D = S x T**

where,

- D ⇒ Distance

S ⇒ Speed

T ⇒ Time

**Relativity**: It is a broad term. It describes about objects moving in either direction or in the same direction, speed of the objects, and time taken by the objects, etc.

The most important thing to remember while solving for the train problems is: do not consider from the front edge of the object and consider from the back edge of the object.

**Example 1**:

A train 100 m long is running at the speed of 30 km/hr. find the time taken by in to pass a man standing near the railway line.

**Solution**:

- Speed of the train = (30 x \(\frac{5}{8}\)) m/sec = (\(\frac{25}{3}\)) m/sec

Distance moved in passing the standing man = 100 m.

Required time taken = (\(\frac{100}{\frac{25}{3}}\)) = (100 x \(\frac{3}{25}\)) sec = 12 sec.

**Example 2**:

A train is moving at a speed of 132 km/hr. If the length of the train is 110 metres, how long will it take to cross a railway platform 165 metre long?

**Solution**:

- Speed of the train = (132 x \(\frac{5}{18}\)) m/sec = (\(\frac{110}{3}\)) m/sec

Distance covered in passing the platform = (110+165) m = 275 m.

Time taken = (275 x \(\frac{3}{110}\)) sec = (\(\frac{15}{2}\)) sec = 7\(\frac{1}{2}\) sec

**Example 1**:

A man is standing on a railway bridge which is 180 m long. He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train and its speed.

**Solution**:

- Let the length of the train be x metres.

Then, the train covers x metres in 8 seconds and (\(x\) + 180) metres in 20 seconds.

∴ \(\frac{x}{8}\) = \(\frac{(x + 180)}{20}\) ⇔ 20\(x\) = 8(\(x\) + 180) ⇔ \(x\) = 120.

∴ Length of the train = 120 m.

Speed of the train = (\(\frac{120}{8}\)) m/sec = m/sec = (15 x \(\frac{18}{5}\)) kmph = 54 kmph.

**Example 2**:

A man sitting in a train which is travelling at 50 kmph observes that a goods train, travelling in opposite direction, takes 9 seconds to pass him. if the goods train is 280 m long, find its speed.

**Solution**:

- Relative speed = (\(\frac{280}{9}\)) m/sec = (\(\frac{280}{9}\) x \(\frac{18}{5}\)) kmph = 112 kmph.

∴ Speed of goods train = (112 – 50) kmph = 62 kmph.

2. \(x\) m/s = \(x\) x \(\frac{18}{5}\) km/hr

3. Time taken by a train of length \(y\) meters to pass a pole or a standing man or a signal post or an object of negligible width would be equal to the time taken by the train to cover \(y\) meters which is primarily the length of the train.

4. Time taken by a train of length \(y\) meters to pass a stationary object of length \(b\) meters is the time taken by the train to cover

(\(y + b\)) meters.

5. Suppose two trains or two bodies are moving in the same direction at \(x\) m/s and \(y\) m/s, where \(x\) > \(y\), then their

Relative speed = (\(x\) – \(y\)) m/s

6. Suppose two trains or two bodies are moving in the opposite direction at \(x\) m/s and \(y\) m/s, then their

Relative speed = (\(x\) + \(y\)) m/s

7. If two trains of length \(a\) meters and \(b\) meters are moving in the opposite direction at \(x\) m/s \(y\) m/s, then the time taken by the faster train to cross the slower train = \(\frac{(a + b)}{(x + y)}\) sec.

8. If two trains of length \(a\) meters and \(b\) meters are moving in the same direction at \(x\) m/s \(y\) m/s, then the time taken by the faster train to cross the slower train = \(\frac{(a + b)}{(x – y)}\) sec.

9. If the two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take \(x\) and \(y\) sec in reaching B and A respectively, then

(A’s speed) : (B’s speed) = \(\sqrt{y}\) : \(\sqrt{x}\)

- Given,

Distance moved in passing the standing man = 200 m

Speed = 20 kmph

Convert kmph to mps

Speed = 20 x \(\frac{5}{18}\) = \(\frac{50}{9}\)

So, Required time = \(\frac{distance}{speed}\)

⇒ Time = \(\frac{200}{\frac{50}{9}}\)

⇒ Time = \(\frac{200 * 9}{50}\)

⇒ Time = 36 sec

Therefore, the time taken by it to pass a man standing near the railway line = 36 sec

**2. A man is standing on a railway bridge which is 200m long. He finds that a train crosses the bridge in 10 seconds but himself in 6 seconds. Find the length of the train and its speed?**

**Solution**:

- Let,

The length of the train = \(x\) m

Then, the train covers \(x\) m in 6 seconds and \(x + 200\) m in 10 seconds.

So, Length of the train is

\(\frac{x}{6}\) = \(\frac{x + 200}{10}\)

⇒ 10 x \(x\) = 6(\(x\) + 200)

⇒ 10 x \(x\) – 6 x \(x\) = 1200

⇒ 4 x \(x\) = 1200

⇒ \(x\) = 300

⇒ Length of train = 300 m

Then, Speed of the train = \(\frac{300}{6}\) m/s = 50 m/s

⇒ 50 x \(\frac{18}{5}\) km/hr

⇒ 180 km/hr

Therefore, length of train = 300m and

Speed of train = 180 km/hr.

**3. A train 320 m long is running with speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the opposite direction in which the train is going?**

**Solution**:

- Given,

Relative Speed of the train and man = (60 + 6) = 66 kmph

Converting it into m/s, i.e.

⇒ 66 x \(\frac{5}{18}\)

⇒ \(\frac{55}{3}\) m/s

Time taken by the train to cross the man is

= time taken by it to cover 320 m at \(\frac{55}{3}\) m/s

= 320 x \(\frac{3}{55}\)

= 17.45 sec

Therefore, time taken by the train to cross the man is 17.45 sec

**4. Two trains 200 m and 320 m long are running in the same direction with speeds of 72 km/hr and 54 km/hr. In how much time will the first train cross the second?**

**Solution**:

- Relative Speed of the trains = (72 – 54) km/hr = 18 km/hr

Convert it into m/sec, i.e.

⇒ 18 x \(\frac{5}{18}\)

⇒ 5 m/s

Time taken by the trains to cross each other is

= time taken to cover (200 + 320)m at 5 m/s

= \(\frac{520}{5}\)

= 104

Therefore, time taken by the trains to cross each other is 104 sec.

**5. A man sitting in a train which is travelling at 60 kmph observes that a goods train, travelling in opposite direction, takes 9 seconds to pass him. If the goods train is 300 m long, find its speed?**

**Solution**:

- Given,

The goods train travels 60 kmph

Relative speed = \(\frac{300}{9}\) m/s

Convert it into kmph, i.e.

\(\frac{300}{9}\) x \(\frac{18}{5}\) = 120kmph

Now, Speed of goods train is

= (120 – 60) kmph

= 60 kmph

Therefore, Speed of goods train = 60 kmph