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Relative motion numerical concept

In Previous post we have study about relative motion continuing that topic we will study some more concept and numerical question here so lets start.

Reference Frame any object motion seen from different location it can be seen different, motion of an object look different from different locations, hence reference frame is that place from which an object motion is seen picture below for reference.

                                     

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.

Thanks for reading 

Relative motion numerical concept

Dated 1st May 2018.