## Train Problems for competition exams

Train Problems is an interesting part of time-distance problems. The Train Problems are a small 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 arise.

Here we will learn certain tricks and see the various forms of train problems.

Train Speed = *Distance*Time

**Formula to convert m/s to Km/hr**

1 meter = 1/1000 km

1 sec = 1/3600 hours

1 m/s = 3600/1000 km/h = 18/5 km/h.

So, **To convert a value in m/s to Km/h, we will multiply it with 18/5.**

**Example:**

20m/s = 20 x *18*5 = *360*5 = 72 km/h

### Some important things you should notice:

1). When two trains are moving in opposite directions their speeds should be added to find the relative speed.

2). When they are moving in the same direction the relative speed is the difference in their speeds.

3). When a train passes a platform it should travel the length equal to the sum of the lengths of train and platform both.

### Some important problems:

**PROBLEM 1.** A 120 m. long train crosses a pole in 27 sec. find the speed of a train in km/h?

**Solution:** By the use of Distance = Speed X Time

**L = S * T **

(where, L= length of train, S= Speed, T= Time).

So,

** L1+L2 = Net speed x T ** (where, Net speed = in same direction S1-S2, And in the opposite direction S1+S2 will be considered as speed. )

So, SPEED= 120/27= 4.44 meter/second

in order to change it in km/hr, we will multiply it with 18/5

4.44 m/s = 4.4x(18/5)=16 km/hrs

**PROBLEM 2.** A 150 m. long train crosses a platform in 45 sec. and crosses a pole in 9 sec. find the length of the platform?

**Solution:** L = S*T …

150 + L2 = S x 45 …..(1)

150 = S x 9……..(2)

By the equation (1) and (2),

150+L2/150 = S x 45/S x 9

150+L2 = 750

Then, L2 = 600 m.