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Mechanics

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 Joined: 29 May 2007 Post: 1
29 May 2007 12:02:35 IST
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relative velocity
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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?

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Joined: 23 Dec 2006
Posts: 374
29 May 2007 13:14:45 IST
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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.

Answer: The direction of the wind is [ 45 - tan -1 (1/5) ] degrees east of north.

Cheers !!

Hot goIITian

Joined: 28 May 2007
Posts: 107
29 May 2007 13:40:58 IST
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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.......

Blazing goIITian

Joined: 19 Feb 2007
Posts: 1802
29 May 2007 13:57:01 IST
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The magnitude of the appearing velocity in both cases has to be the same right. Only the direction differs?

Hot goIITian

Joined: 28 May 2007
Posts: 107
29 May 2007 14:06:02 IST
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i don't think it is so..

Blazing goIITian

Joined: 19 Feb 2007
Posts: 1802
29 May 2007 14:07:42 IST
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if not then there are too many unknowns here, there is something missing in the question i guess. (Thats what i think...i might be wrong)

Hot goIITian

Joined: 28 May 2007
Posts: 107
29 May 2007 14:09:29 IST
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Forum Expert
Joined: 13 Nov 2006
Posts: 558
28 Jun 2007 10:27:40 IST
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let the man velocity be v = u i + u j in vector form.
Let the wind velocity be w = x i + y j.

There fore the relative velocity = w - v = (x - u) i + (y - u) j
But the direction of the relative velocity is from north. Therefore x - u = 0. or x = u.

Now when the man velocity is doubled 2v = 2u i + 2u j.
therefore w - v = relative velocity.

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Joined: 10 Apr 2007
Posts: 1777
28 Jun 2007 10:44:41 IST
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Wonderful solution Rammana sir. Hats off.

Forum Expert
Joined: 13 Nov 2006
Posts: 558
28 Jun 2007 11:04:54 IST
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Now   (x - 2u) / (y - 2u) = 2 .
Solve for the requirement.

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