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

- D ⇒ Distance
S ⇒ Speed
T ⇒ Time

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

Distance moved in passing the standing man = 100 m.

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

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

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

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

- Let the length of the train be x metres.

Then, the train covers x metres in 8 seconds and ([latex]x[/latex] + 180) metres in 20 seconds.

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

∴ Length of the train = 120 m.

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

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

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

2. [latex]x[/latex] m/s = [latex]x[/latex] x [latex]\frac{18}{5}[/latex] km/hr

3. Time taken by a train of length [latex]y[/latex] 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 [latex]y[/latex] meters which is primarily the length of the train.

4. Time taken by a train of length [latex]y[/latex] meters to pass a stationary object of length [latex]b[/latex] meters is the time taken by the train to cover ([latex]y + b[/latex]) meters.

5. Suppose two trains or two bodies are moving in the same direction at [latex]x[/latex] m/s and [latex]y[/latex] m/s, where [latex]x[/latex] > [latex]y[/latex], then their Relative speed = ([latex]x[/latex] - [latex]y[/latex]) m/s

6. Suppose two trains or two bodies are moving in the opposite direction at [latex]x[/latex] m/s and [latex]y[/latex] m/s, then their Relative speed = ([latex]x[/latex] + [latex]y[/latex]) m/s

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

8. If two trains of length [latex]a[/latex] meters and [latex]b[/latex] meters are moving in the same direction at [latex]x[/latex] m/s [latex]y[/latex] m/s, then the time taken by the faster train to cross the slower train = [latex]\frac{(a + b)}{(x - y)}[/latex] 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 [latex]x[/latex] and [latex]y[/latex] sec in reaching B and A respectively, then (A's speed) : (B's speed) = [latex]\sqrt{y}[/latex] : [latex]\sqrt{x}[/latex]

- Given,

Distance moved in passing the standing man = 200 m

Speed = 20 kmph

Convert kmph to mps

Speed = 20 x [latex]\frac{5}{18}[/latex] = [latex]\frac{50}{9}[/latex]

So, Required time = [latex]\frac{distance}{speed}[/latex]

⇒ Time = [latex]\frac{200}{\frac{50}{9}}[/latex]

⇒ Time = [latex]\frac{200 * 9}{50}[/latex]

⇒ Time = 36 sec

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

- Let,

The length of the train = [latex]x[/latex] m

Then, the train covers [latex]x[/latex] m in 6 seconds and [latex]x + 200[/latex] m in 10 seconds.

So, Length of the train is

[latex]\frac{x}{6}[/latex] = [latex]\frac{x + 200}{10}[/latex]

⇒ 10 x [latex]x[/latex] = 6([latex]x[/latex] + 200)

⇒ 10 x [latex]x[/latex] - 6 x [latex]x[/latex] = 1200

⇒ 4 x [latex]x[/latex] = 1200

⇒ [latex]x[/latex] = 300

⇒ Length of train = 300 m

Then, Speed of the train = [latex]\frac{300}{6}[/latex] m/s = 50 m/s

⇒ 50 x [latex]\frac{18}{5}[/latex] km/hr

⇒ 180 km/hr

Therefore, length of train = 300m and

Speed of train = 180 km/hr.

- Given,

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

Converting it into m/s, i.e.

⇒ 66 x [latex]\frac{5}{18}[/latex]

⇒ [latex]\frac{55}{3}[/latex] m/s

Time taken by the train to cross the man is

= time taken by it to cover 320 m at [latex]\frac{55}{3}[/latex] m/s

= 320 x [latex]\frac{3}{55}[/latex]

= 17.45 sec

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

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

Convert it into m/sec, i.e.

⇒ 18 x [latex]\frac{5}{18}[/latex]

⇒ 5 m/s

Time taken by the trains to cross each other is

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

= [latex]\frac{520}{5}[/latex]

= 104

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

- Given,

The goods train travels 60 kmph

Relative speed = [latex]\frac{300}{9}[/latex] m/s

Convert it into kmph, i.e.

[latex]\frac{300}{9}[/latex] x [latex]\frac{18}{5}[/latex] = 120kmph

Now, Speed of goods train is

= (120 - 60) kmph

= 60 kmph

Therefore, Speed of goods train = 60 kmph