A MAN IS MOVING IN N-E DIRECTION AND WIND APPEARS TO BE BLOWING FROM NORTH. WHN MAN DOUBLES HIS SPEED WIND APPEARS INTHE DIRECTION tan^-12 EAST OF NORTH.WHAT IS THE ACTUAL DIRECTION OF WIND?

Sorry, my computer cannot upload the image. But I will try to give the solution in full details.

 

The resultant R of two vectors P & Q ( the angle between the two vectors P & Q is   ) makes an angle  with P, such that,

 

tan  = Q sin / ( P + Q cos )

 

Here, we shall apply the above formula.

 

Let the,

velocity of the man = v in the N.E direction.

velocity of the wind = w

the relative velocity of the man with respect to the wind = v - w = v + ( - w )

To the man, the wind appears to blow from the north i.e the direction of the resultant vector of v and - w (i.e the relative velocity of the man with respect to the wind ) is in the south direction.

 

Let the angle between the velocity of the man v and the vector - w be

 

Now, it is clear so far that since the vector v is in N.E direction and the vector sum of v and - w is the south direction then - w must be somewhere in the S-W direction.

 

So, now, framing the first equation,

 

tan 135 = w sin  / ( v + w cos  )

 

( since, here the angle between the vector v and the resultant of v & ( - w )  is 135 degrees )

 

i.e - 1 = sin  / ( f + cos  ) where, f = v / w.........eqn. (1)

 

Now when the speed of the man gets doubled the resultant velocity ( i.e the relative velocity of the man with respect to the wind ) shifts towards the vector whose magnitude has increased i.e towards the velocity of the man and now the resultant makes an angle of tan^-12 east of north i.e the angle between the resultant and the 2v vector is (tan^-12 - 45) degrees.

 

Now, we know, tan(x-y) = (tanx - tany) / (1+tanx.tany)

So, tan (tan^-12 - 45) = ( tan tan^-12  - tan 45 )/( 1 +tan^-12 . tan 45 )

                             = (2-1) / (1+2.1) = 1 / 3

 

Now, let us form the second equation,

 

1/3 = w sin  / ( 2 v + w cos  ) = sin  ( 2 f + cos  )...........eqn. (2)

 

Those of you who wonder why I have taking f = v / w, then let me tell you in the above two equations we have three unknows v, w and , so I've converted the two unknowns v,w into one unknown f = v / w. Now we have two unknows f &  and just two equations.

 

Eliminating f from the above two equations,  we have,

 

- cos  = 5 sin

 

i.e tan  = - 1 / 5

 

Now, as estimated earlier that reverse velocity of the wind is in the S-W direction & the velocity of the man is in the N.E direction, so it is obvious that ,

 = 180 + tan ^-1 ( - 1 / 5 ) degrees = 180 - tan ^-1 (1/5) degrees

 

Now, so the reverse velocity of the wind is ( 180 - tan ^-1 (1/5) - 135 ) degrees west of south, i.e  [ 45 - tan ^-1 (1/5) ] degrees west of south.  

 

The direction of the vector - w is [ 45 - tan ^-1 (1/5) ] degrees west of south.

 

So, the direction of vector w is [ 45 - tan ^-1 (1/5) ] degrees east of north.

 

 

Cheers !!

let the speed of the man b 'p' and of the wind be 'q' then and $ be the angle between the direction of the wind and the direction of the man
tan 45=qsin$/p-qcos$=1===>q sin$=p-q cos$--------1

now the speed of the man is doubled ==> speed is '2p'
the angle of the resultant =arc tan 2 eastof north ==> angle between the direction of the man and resultant=arc tan2 -arc tan 1=q sin(180 -$)/2p-q cos(180-$)=1/3 [since the angle arc tan 2 is greater than 45 the resultant comes nearer to east..]===>3q sin$=2p+q cos$-----2
from 1 and 2,
3qsin$-q cos$=2q(sin$+cos$)
3 sin$-cos$=2sin$+2cos$
==>sin$=3 cos$
==>tan$=3
==>$=arc tan 3 [from the direction of the man towards north..] or arc tan 2 west of north...

 

plzzzz.. correct me if m wrong.......

Now   (x - 2u) / (y - 2u) = 2 .

Solve for the requirement.