Ship' smoke velocity? I am studying mechanics from MIT 2.003. In HW problem "which I solved wrong" they stated that:

An important fact is that the smoke travels with the wind at the velocity of the wind


In this problem ship is travelling with a constant speed.
Questions:


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*Why smoke travels with the wind at the velocity of the wind ? shouldn't it be relative to the ship?


Thanks in Advance.
Additional edit: the main question as asked in the exercise in reference here is:

The ship in the diagram travels at a constant speed of $V_{\text{s}}=20~\mathrm{m/s}$. The wind is blowing at a speed of $V_{\text{w}}=10~\mathrm{m/s}$, with directions as shown in the figure. Determine the magnitude and direction of the smoke coming from the smokestack as it appears to a passenger on the ship. It is strongly recommended that you use vectors to solve this problem, so as to warm up for more difficult vector-based problems soon to follow. 

 A: The simple answer is that the smoke does have the velocity of the boat the moment  it leaves the smoke stack, but very quickly is accelerated by drag to match wind speed because of its negligible inertia.
As is, the question is poorly worded. It should include a statement about this assumption (e.g. "don't ignore air resistance" or "assume the smoke has negligible inertia") since many physics problem at the introductory level do ignore air resistance. 
A: You might want to think of the smoke as a short series of puffs of smoke emitted from the stack. Each subsequent puff originated in a different location because the ship is moving; but once it leaves the stack it travels at the velocity of the wind.
I recommend that you draw the boat at two time points - say one second apart - and consider where the first puff is once the second puff is emitted. That diagram will tell you everything you need to solve this.
A: Note that the ship is assumed to not give the smoke a horizontal velocity component. This is an assumption in this question. Consequent to this assumption, if there were no wind, but the ship was moving, the smoke would move away from the ship in a line showing the previous positions of the ship, i.e. the smoke would move in the direction $-V_\rm s$. If there was wind, and the ship was not moving, the smoke would move away in the direction of wind motion $V_\rm w$.
To solve the problem, you can use a drawing at a scale of $1mm = 1m/s$. Using the notation in your diagram above (a single letter means with reference to the earth): start at point $O$, draw the vector (line at an angle) of $- \vec V_\rm s$, then starting from the end position of this $- \vec V_\rm s$, draw the vector of $\vec V_\rm w$. Now, connect point $O$ to the end of your $\vec V_\rm w$. The length and direction of this last connecting line are the velocity 
 and direction of the smoke as seen by a passenger on the ship. In other words:
$\vec V_\rm{w/s} = -\vec V_\rm s + \vec V_\rm w \tag1$
If you are not into accurate drawing, in a rough sketch, use the $1mm = 1m/s$ scale as mentioned above, and then determine the magnitude of the $x$ and $y$ vector components for $-\vec V_\rm s$ and $\vec V_\rm w$ using trigonometry ($sin\theta$, $cos\theta$ and the property $x^2 + y^2 = z^2$ for a triangle with hypotenuse $z$, and $x$ and $y$ as the two other sides). Add the $x$ components and the $y$ components of $- \vec V_\rm s$ and $\vec V_\rm w$ to give you a new vector endpoint. This is equal to $\vec V_\rm{w/s}$.
