d Formula Technique 14 & 15 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

# Formula Technique 14 & 15 Formula Technique 14: Train(s) Crossing Objects and Moving in Same / Opposite Direction

( basic maths formulas pdf, basic mathematical formulas pdf free download )

In questions about train(s), we should remember that a train has certain length, and we will have to take into consideration its length to solve such problems; If this train completely crosses a pole or a point at ‘A’, then it will have to travel a distance equal to its length.

Example: If a train whose length is “200 meters crosses an electric pole at point A with a speed of  50 m/s. In what time it will completely cross the electric pole?

Solution:

Of course, the train will completely cross a pole only if its locomotive power to it last boggy all go ahead the pole! i-e first locomotive will cross the pole, then first boggy will cross it, then second and so on till the last boggy. In this way to cross the pole, the whole train will cover a distance equal to its length.

So, Distance traveled =s = length of train = 200 m Speed = v = 50 m/s

Time to cross the pole = t = ?

We use relation;

S=vt

or  t=S/v

or t = 200/50

or t= 4 sec

o the train will pass or cross the pole in 4 seconds!

Special Cases

1- When a train crosses across a bridge, a platform, a play fields  etc, it covers a distance equal to its length plus the length of bridge, platform, or play ground etc. 2- When two trains moving in opposite directions, then while crossing each other they will both travel distance equal to sum of lengths of both trains 3- When two trains moving in the same direction, Then net speed is the difference of  speeds of both trains.

4- When two trains are moving in opposite directions , then net speed is the sum of speeds of both trains.

Formula Technique 15: A Boat In A River

This is another situation of boat sailing on water.

1- Case One (speed down)

When boat is sailing in same direction as of water flow , then;

Total Speed down of boat = Speed of boat + Speed of flowing Water

2- Case Two(speed up)

When boat is sailing in opposite direction of flow of water, then

Net speed up of boat = Speed of boat – Speed of flowing  water