Problems on trains are very common questions asked in various competitive examinations like – SSC, Bank PO, Clerk etc. You can solve any questions on trains using the concept of time, speed and distance. We use formulas related to time, speed and distance to solve the problems on trains.

Important points to remember while solving problems on trains

A. When a train crosses a stationary object (e.g. man, lamp post, pole, sign), then, speed of object is 0. I mean it does not move. So, You have to only bother about speed of train. Also, distance covered in this situation is length of the train. Practically speaking, every object has some length. But, This is negligible in comparison with length of the train. So, we consider length of object as 0. So, total length in this situation is length of the train.

B. When a train crosses a platform/bridge etc, then, in this situation, speed of platform/bridge is 0 i.e. it is stationary. So, you have to bother about speed of train only. But, there is some length of platform/bridge etc. that are not negligible in comparison with the length of train. So, total distance covered is combined length of platform/bridge and length of train. The same condition applies when the train crosses anything that has a considerable length.

C. When two trains are moving in the same direction, then, their relative speed is the difference between the speeds of trains. In this situation, total distance covered is combined length of both trains.

D. When two trains are moving in opposite direction, then, their relative speed is the sum of the speed of trains. In this situation, total distance covered is combined length of both trains.

Important formulas to remember while solving problems on trains

A. When a train A1, having length L1 and running with a speed of u1 km/h, crosses a stationary object (e.g. man, lamp post, pole, sign) that has negligible length in comparison with length of the train. Then,

Total distance covered by train = L1 (Length of the train).
Time taken to cross the object = $( \frac {L1}{u1} )$ hour

B. When a train A1, having length L1 and running with a speed of u1 km/h, crosses a platform/bridge that has some significant length, L2, with the length of the train. Then,

Speed of platform/bridge = 0
Speed of train = u1 km / h
Total distance covered by train = (L1 + L2) km.
Total time taken to cross the object (platform/bridge) = $( \frac {L1 + L2}{u1} )$ hour

C. When two trains (A1 and A2) of length L1 and L2 are moving in the same direction with speed u1 km/h and u2 km/h. Then,

If u1 > u2, Relative speed = (u1 – u2) km / h
Train A1 will cross train A2. Total distance covered is combined length of both trains. So, total distance will be L1 + L2. Total time taken is $( \frac {L1 + L2}{u1 – u2} )$ .

If u2 > u1, then, train A2 will cross train A1. Formulas is similar as above. Just interchange values of train A1 and A2.

If u2 = u2, then, both train will run together. One train will never cross another.

D. When two trains (A1 and A2) of length L1 and L2 are moving in the opposite direction with speed u1 km/h and u2 km/h. Then,

Relative speed = (u1 + u2) km / h
One train will cross another and total distance covered is combined length of both trains. So, total distance will be L1 + L2. Total time taken is $( \frac {L1 + L2}{u1 + u2} )$ hour.

Some examples to demonstrate on how to solve problems on trains

1. What is the time taken to cross a man by a train of length 200 metres running with a speed of 10km/h.
Solutions –
Total distance covered = 200 metres.
Speed of the train = 10 km/h = $ 10 \times { (\frac {5}{18}) } = ( \frac {25}{9} ) $ metres per second.
Time taken = Total distance covered / speed of the train = $ \frac {200}{(\frac{25}{9})} = \frac {(200 \times 9)}{ 25} = 72 $ secs.

2. Find the speed of the train, having length 200 metres, that crosses a 100 metres long platform in 10 secs.
Solutions –
Total distance covered to cross the platform = 200 + 100 = 300 metres.
Time taken to cross = 10 secs.
Speed of train = $( \frac {Total Distance covered}{Total Time Taken} )$
=> $ \frac {300}{10}$
=> 30 m/s

[latexpage]