IIT JEE , AIEEE Forum - Physics , Mechanics

biki

Two persons A and B are running along the same direction and they see a car moving. A says that the car is moving along EAST and B says that the car is moving along NORTH but both say that the speed of the car is same. If velocity of B is twice that of A , find the true direction of motion of the car...

i m extermely sorry as i had mention west whereas it is actually north......???

spideyunlimited

Let speed of A = x m/sec
Speed of B = 2x m/sec
Let actual speed of car = S

Take the direction of motion of A and B as positive.

Now, if the motion of car appears to be along different directions means its speed is less than the greater one and more than the lower one.

So speed of Car lies between A and B and is along same direction.

It is moving towards east, and A and B are too. As A's speed is less than Car, the car appears to be moving East but as B's speed is more than the Car, the Car appears to be moving in the opp. direction ie. West

Acc. to the given information
Relative speed of B wrt car = Relative speed of car wrt A
S - 2x = x - S ...(Or the other way around by multiplying both sides with -1...one speed is greater than car's and one is less that is why direction of motion of car seems different)

S - 2x = x - S
2S = 3x
S = 3x/2 m/sec towards east

*PLZ RATE*

biki

hey...spidey....

i m sorry man..... B sees it towards north.....

but i m saluting you for your efforts..... you are absolutely correct....

PLEASE SOLVE IT NOW................. !!!!!!!

coolank2

Well, I think there will not be any particular (i.e. standard) direction of that car.

Let i and j be the unit vectors along E-N direction.

Say, vel of A = ai + bj ;

vel of B = mi + nj ;

vel of car = xi + yj

Now, A and B are moving in same direction and vel of B is twice that of A .....

i.e. m = 2a and n = 2b ..... as the direction remains same

Now v_car/man = v_car/ground - v_man/ground

Thus for A: ( )i + 0j = (x-a)i + (y-b)j ..... i.e. y = b

Thus for B: 0i + ( )j = (x-m)i + (y-n)j ..... i.e. x = m

I've left the spaces blank .... as we know only the directions of relative motion ....

Thus, vel of car = mi + bj = 2ai + bj

And dir is tan^-1(b/2a) .....

Let me know if I'm wrong ....

biki

the options are.............

1) tan^-1 ( 2 ) along north of east
2) tan^-1 (1/2) along north of east
3) tan^-1 (2) along south of east
4) none

jasmeetawal

well the answer is tan^-11/2 along north of east

assume the velocity of 'A' be i₁ + j₁ and of car be i₂+j₂.

then the velocity of 'B' will be 2i₁ + 2j_1.

relative velocity of car with respect to 'A'

i₂+j₂-(i₁+ j₁)=i

as its relative velocity is towards west

j₂- j₁=0,

j₂=j₁. and i₂ - i₁=i.....(1)

now relative velocity of car with respect to 'B'

2i₁+ 2j₁- (i₂+j₂)=j

2i₁ - i₂=0

i₁=i₂/2

2j₁-j₂=j

but j₁=j₂

j₂=j

putting the value of i₁ in (1).

i₂-i₂/2=i

i₂/2=i

but magnitude of i = j.so,

i₂/2=j₂

j₂/i₂=1/2

tan# =1/2

direction of car is along tan^-11/2 north of east.

biki

its north and east.....

north for B and east for A

jasmeetawal

thats what i have assumed.
for b it is north so relative velocity of car with respect to 'B'
will only have north component i.e. j.
and for 'A' it will just have east component i.e. i.

mastershubham

great...jasmeetawal

you are absolutely correct.

biki

ya i got it now...

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