Problems based on train is a part of Time and Distance chapter but moving and stationary objects make this chapter somewhat different. Today I am going to help you revise this chapter using simple examples.

### Formula:

#### Some useful concepts :-

**I.**To change

**k/h in m/s**we multiply by

**5/18**

E.g. 72 k/h means

72 × 5/18 = 4 × 5 = 20 m/s

**II.**To change

**m/sec**in k/h we multiply by

**18/5**

E.g. 45m/s means

45 × 18/5 = (9 × 18) k/h = 162 k/h

**III.**In the concept of trains, there are two objects. First object train and second is that which is crossed by the train.

**Tips:**

**When a moving train crosses a man, the first object is train and second object is man.**

__Case 1 –__**When a moving train crosses a platform, first object is train and second is platform.**

__Case 2 –__

**When a moving train crosses a man who is standing on a railway platform, first object is train and second is man.**

__Case 3 –__**When a moving train crosses another moving train (same or opposite) direction; the first object i s first train and second is second train.**

__Case 4 –__**When a moving train crosses a man who is standing in another moving train, the first object is first train and second one is man (not 2**

__Case 5 –__^{nd}train)

__Rule I:__

**When a train crosses a man/pole/tree (in stationary form)**

Let a train is having a speed is S and the length L, crosses a pole in T time then,

Here the breadth of man, tree, pole is negligible with respect to the train.

__Rule II:__

__When train crosses a platform__.

Let a train having speed of S and length L, crosses a platform whose length is L2, in time T, then

__Rule III :__

**When a train crosses another running train**

Let a train having Speed S, and length L, crosses another train which is travelling in opposite direction at a speed of S2 and L2 in time T. Then-

Both object having length and speed.

__NOTE:__

**(i)**If both object having length (L1 and L2) then, the Length is always sum i.e. (L1 and L2)

**(ii)**If both object having speed S1 and S2, then the speed are added. (When direction is opposite) and subtracted (When direction is same)

__Rule IV:__

**When a moving train crosses a man who is running**

Let a train having a speed S1 and length L, crosses a man, who is running at a speed S2 in the same direction T time, then

Since man has negligible breadth

__Rule V:__

**When a moving train crosses a man who is sitting into a moving train.**

Let a train having a speed S1 and length L1 crosses a man who is sitting in another train which is travelling in opposite directions at a speed of S2 and L2 in time T

EXAMPLES:

__1)__**Find the time taken by a train 180m long, running at 72 km/h in crossing an electrical pole.**

__Solution –__

( Because pole has no breadth)

__2)__**A train 140m is running at 60 km/h. In how much time will it pass a platform 260m long?**

**Solution:**

__3)__**A train 100m long is running at a speed of 70 km/h. In what time will it pass a man who is running at a speed of 10 km/h in the same direction?**

**Solution:**

__4)__**A train 160m long taken 30 sec in crossing a tunnel 440m long. The speed of the train is**

**Solution:**

__5)__**A man is standing on a railway bridge which is 50m long. He finds that a train is crossed on the bridge in 4 ½ sec but himself in 2 sec. Find the length and speed of train**

__Solution:__From 1 and 2

4.5 L = 2 L + 100

2.5 L = 100 = L =40m

S= 40/2

**= 20m/s**

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