Well relative motion is depend upon frame of reference. So what is frame of reference ? . It is a reference point from which measurement is done. Hence there are two types of frame of reference. Which are called inertial and non inertial frame of reference . Therefore inertial frame of reference is stationary . And non inertial frame is moving frame of reference .

If you want to know more about frame of reference. Then visit my previous post about frame of reference. Where details about frame of reference is given. So i am sharing that link below in this post.

In this post, i am going to share numerical concept for relative motion. How you are solving numerical problems ? And what are concept involved for numerical questions ?

So before going to solve numerical problem. There are some important point to keep in mind. Because this point is helping to solve numerical problem easily.

Relative motion definition is dependent on reference frame. Because without reference frame you are not able to define relative motion. Hence for relative motion, you always need a reference frame. But you are thinking why reference frame is required ?

Well this is because of nothing is at rest in the universe. All particles and every body are in the motion. So without reference frame, it is very difficult to find out motion of any body. Therefore reference frame is must required. So lets talk about reference frame.

Hence** Reference Frame **is define as any object motion, seen from different location. And it is found that motion of an object looks, different from different locations. Therefore reference frame is that place, from which an object motion is seen. So refer the picture given below for reference frame.

How same object motion is looking different from different reference frame. So ground standing person is observing aeroplane motion different. And in the car person observe different motion for the same aeroplane.

Relative motion numerical concept |

Here the person sitting inside airplane in front sit coordinate are not changing with respect to airplane but with respect to man observing from ground their coordinate are changing.hence ground man reference frame is ground and airplane passenger reference frame is airplane now we will see some numerical question.

Q two trains of length 400 m each are running on two parallel track with uniform speed of 72 km/hr in the same direction A ahead of B.the driver of B decided to overtake A and accelerate by 1 m/s² if after 50 sec the guard of B just brush pass the driver of A what is original distance between them.

Ans this type of question may be long but it is simple.

first as per question it is clear that train A is ahead.

B = 400 m A = 400 m

→→→→→→→ →→→→→→→

Ub = 72 km/h ← d → Ua = 72 km/hr

now here first we will calculate speed in m/s so Ua = 72*5/18 = 20 m/s = Ua = Ub = 20 m/s

now from question given B accelerate with 1 m/s² hence one time will come both A and B will be parallel to each other and again after that B will overtake A at this point Guard of B and driver of A will be in front at present A is ahead.

from question given that at 50 sec B crosses the A now to Cross A total distance travelled by B will be d+400+400 = d+800 m now total time take 50 s now both are moving in same direction so take relative velocity hence initial relative velocity = 20-20 =0 its mean that if both trains travel with this speed then B will not overtake but her driver B accelerate with 1 m/s² now relative acceleration Aa – Ba = 1-0 = 1 m/s² now we have data.

s = d+800, u = 0, a = 1 , t = 50 .

s = ut+1/2at² hence put all value in this equation.

d+800 = 0*50+1/2*1*50*50

d+800 = 1250 or d = 450 m this distance is initial separation between two train but in our question distance between Guard of B and Driver of A it will be d+400+400 = 450+800 = 1250.

Q Two car are moving with speed of 30 m/s and 40 m/s in opposite direction with uniform velocity initial distance between the is 1200 m at what time they will meet each other and at what distance from car A suppose length of car is negligible.

Ans from question both car are moving in opposite direction .

A B

→ d ⊗ ←

Ua = 30 m/s Ub = 40 m/s

Now suppose red cross is meeting point of both cars now suppose from car A to d distance both car meet.

we can write d = Uat or d = 30t .

Now distance between this two is 1200 m so use formula

s = ut +1/2at² here a = 0

1200 = (30+40)t or t = 1200/70 = 17.14 s

d = 30*17.14 = 514.2 m

Many more question you can do yourself after understanding concepts.

If you want to know more about relative motion. Then refer my previous post. This link is given below.

I hope you are enjoying learning relative motion numerical concept. If you are having any doubt, feel free to ask any question. Because my motive is to solve your problem. So i am here to help you always. Thanks for sharing in social media, comment and like. Learn and grow.

Dated 1st May 2018.

I'm Telecom Engineer by profession and Blogger by passion