Train Problems form an interesting portion of the time-distance problems. The Train Problems are a bit different than the regular problems on the motion of the objects. This is due to the finite size of the trains. As a result of the length of the trains, many interesting train problems originate. Here we will learn certain tricks and see the various forms of the train problems.

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## Train Problems

Due to the small size of the cars and other objects, we take them as point objects. Since the size of the trains is comparable to the distances that they may travel, then we will have to take the size or the length of the trains into account too. The same formulae that we saw already are applicable here too. Here we shall see some examples of this concept and then learn some tricks from other examples.

Example 1: A train running at the speed of 60 km/hr. It crosses a certain pole that is in the way in 9 seconds. What is the length of the train?

A) 200 meters B) 180 meters C) 376 meters D) 150 meters

Answer: The train crosses the pole in 9 seconds. This means that from the point when the engine or the front of the train crosses the pole to the point when the back of the train reaches the pole, we have 9 seconds. Thus we can get the length of the train by calculating the distance that the train travels in 9 seconds.

We have the speed of the train = 60 km/hr. We have to convert it into m/s. Let us see the trick to convert km/hr to m/s:

1km/1h = 1×(1000/60×60) = 1×(5/18) m/s.

Therefore if the speed is ‘x’ km/hr then we can change it to m/s by multiplying x with 5/18. Let us do it for the current example.

We have the speed of the train = 60 km/hr or 60×(5/18) = 50/3 m/s.

Therefore in 9 seconds, the train will cover a distance = speed×time

In other words, we have the Distance = (50/3 m/s)×9 = 150 m. Therefore the correct option is D) 150 meters. Let us see some more examples that can be formulated on the basis of this concept.

**Browse more Topics under Time And Speed**

- Distance/Speed Relation
- Data Sufficiency
- Relative Speed and Conversions
- Time & Speed Practice Questions

## Solved Examples

Example 2: A certain train is 125 m long. It passes a man, running at 5 km/hr in the same direction in which the train is going. It takes the train 10 seconds to cross the man completely. Then the speed of the train is:

A) 60 km/h B) 66 km/h C) 50 km/h D) 55 km/h

Answer: Here we will have to use the concept of the relative speed. The relative speed of two objects is the sum of their individual speeds if they are moving opposite to each other. If the two objects are moving in the same direction, then their relative speed is equal to the difference between the two speeds. Hence if the man was at rest or we can with respect to the man = ( 125/10 )m/sec

Therefore, this speed of the train = ( 25/2 )m/sec.

In other words we can write, this speed = ( (25/2) × (18/5) ) km/hr = 45 km/hr.

Now, let the speed of the train be = x km/hr. Then, the relative speed of the train with respect to the man = (x – 5) km/hr.

Therefore, we must have, (x – 5) = 45 or x = 50 km/hr. Therefore, the correct option is C) 50 km/h.

Example 3: Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:

A) 2: 3 B) 3:2 C) 1:3 D) 3:1

Answer: Let the speeds of the two trains be = x m/s and y m/s respectively. Then, the length of the first train = 27x meters, and length of the second train = 17y meters.

We can write:

( 27x + 17y ) / (x+ y )= 23

Or 27x + 17y = 23x + 23y, therefore we have: 4x = 6y

and ( x/y ) = ( 3/2 ). Hence the correct option here is B) 3:2

Example 4: A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?

A) 20 s B) 30 s C) 40 s D) 50 s

Answer: We have already seen the formula for converting from km/hr to m/s: x km/hr =[ x × ( 5/18) ]m/s.

Therefore, Speed =[45 x (5/18) m/sec = (25/2)m/sec.

Thus the total distance to be covered = (360 + 140) m = 500 m. Also, we know that the formula for finding Time = ( Distance/Speed )

Hence, the required time = [ (500 x 2)/25 ] sec = 40 sec.

## Practice Questions

Q 1: A jogger running at 9 km/h alongside a railway track in 240 meters ahead of the engine of a 120 meters long train running at 45 km/h in the same direction. In how much time will the train pass the jogger?

A) 66 s B) 72 s C) 63 s D) 36 s

Ans: D) 36 s

Q 2: Two trains are moving in opposite directions at a speed of 60 km/hr and 90 km/hr respectively. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:

A) 44 s B) 48 s C) 52 s D) 200 s

Ans: B) 48 s

Q 3: Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:

A) 25 m B) 50 m C) 75 m D) 100 m

Ans: B) 50 m

How can u solve ex 4

LIterally contradict yourself by saying distance is greater than or equal to displacement then saying that displacement can never be less than distance but if distance is greater than or equal to then displacement would be less than or equal too sick contradiction

For knowing more about speed and velocity click the following link

Many of your questions are wrong let alone solutions. These waste our precious time and cause confusion

The answer for Example 4 is wrong. The correct answer 425.6.

yeet