# Train Problems

#### Chapter 19

5 Steps - 3 Clicks

# Train Problems

### Introduction

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

### Methods

Conversion: It includes conversion of kilometer per hour (kmph) into meter per second (mps) or vice -versa.

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.

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.

### Formulae

1. $$x$$ km/hr = $$x$$ x $$\frac{5}{18}$$ m/s

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) = $$\sqrt{y}$$ : $$\sqrt{x}$$

### Samples

1. A train of length 200 m is running at the speed of 20 km/hr. Find the time taken by the train to pass a man standing near the railway line?

Solution:

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