Wind Velocity: Findings & Insight Needed

[Iniuria12]

Having a little bit of trouble with this and not sure if my findings\thoughts are right and would like some insight.

If a boat is traveling south at 10 km/h and a person on the deck measures the wind velocity to be 8 km/h east, would the actual wind velocity (velocity of wind relative to the ground) still be 8 km/h east or would the answer be resolved by drawing a vector diagram and having a velocity roughly 6 km/h at heading 53 degrees east of south??

Haven't been able to find a definitive answer in correlation to this problem and not sure if I'm over-examining the question or not.

Appreciate any insight.
Thanks

 

[learningphysics]

Having a little bit of trouble with this and not sure if my findings\thoughts are right and would like some insight.

If a boat is traveling south at 10 km/h and a person on the deck measures the wind velocity to be 8 km/h east, would the actual wind velocity (velocity of wind relative to the ground) still be 8 km/h east or would the answer be resolved by drawing a vector diagram and having a velocity roughly 6 km/h at heading 53 degrees east of south??

Haven't been able to find a definitive answer in correlation to this problem and not sure if I'm over-examining the question or not.

Appreciate any insight.
Thanks

It depends on how the question is phrased... if the wind is 8km/h east relative to the boat, then the wind velocity relative to the ground is different. I think in your example, since they explicity said that someone on the deck measured the wind velocity, they mean to say that the 8km/h is relative to the boat and not to the ground... But I don't get 6km/h... how did you get that?

Suppose two objects are moving... x is the velocity of object 1 relative to the ground (or any reference)... y is the velocity of object 2... (note that x and y are vectors)

the velocity of y relative to x is y-x. The velocity of x relative to y is x-y.

 

[Iniuria12]

Ok.. So now I'm royally confused. I presume I'm just working myself up over nothing and just confusing which calculation method I should be using.

Drawing a vector diagram, I would come up with results, boat:10 km/h south, wind:8 km/h east, and resultant:13 km/h, 39 degrees, east of south..

Now with the velocity of y relative to x being y-x, that would be 8 km/h - 10 km/h = -2 km/h.

In other words, I'm just really unsure as to what I would use as my absolute answer, thus being actual wind velocity relative to the ground.

 

[learningphysics]

Ok.. So now I'm royally confused. I presume I'm just working myself up over nothing and just confusing which calculation method I should be using.

Drawing a vector diagram, I would come up with results, boat:10 km/h south, wind:8 km/h east, and resultant:13 km/h, 39 degrees, east of south..

Now with the velocity of y relative to x being y-x, that would be 8 km/h - 10 km/h = -2 km/h.

In other words, I'm just really unsure as to what I would use as my absolute answer, thus being actual wind velocity relative to the ground.

Well... this is how I understand the problem... You're given the velocity of the boat relative to the ground... I'll call it $$\overrightarrow{v_b}$$ = 10km/h south. You're also given the velocity of the wind relative to the boat. I'll call this I'll call it $$\overrightarrow{v_{wb}}$$ = 8km/h east.

And what you want to calculate the velocity of the wind relative to the ground... I'll just call that $$\overrightarrow{v_w}$$.

We know that $$\overrightarrow{v_{wb}}$$= $$\overrightarrow{v_{w}}$$ - $$\overrightarrow{v_b}$$ (I hope this makes sense, let me know if it doesn't).

So $$\overrightarrow{v_w}$$ = $$\overrightarrow{v_{wb}}$$ + $$\overrightarrow{v_b}$$

So calculating the above resultant using vector geometry should give your answer. Actually that's exactly what you calculated... So your answer 13 km/h, 39 degrees, east of south is the velocity of the wind relative to the ground...

Remember that these are vectors so when adding or subtracting you can't just deal with the magnitudes... So to calculate $$\overrightarrow{x}$$ - $$\overrightarrow{y}$$, you would be calculate this resultant:
$$\overrightarrow{x}$$ + (-$$\overrightarrow{y}$$) using vector geometry.

Did you post the question exactly as it is written? The question asks to find the velocity of the wind relative to the ground?

 

[Iniuria12]

Question: Boat traveling south at 10 km/h. A seaman on deck measure wind velocity to be 8 km/h east. What is the actual wind velocity? (the velocity of the wind relative to the ground)

Basically your statement $$\overrightarrow{v_{wb}}$$ = $$\overrightarrow{v_{w}}$$ - $$\overrightarrow{v_b}$$ would be the same as saying True Wind Speed = Apparant Wind Speed + Boat Wind Speed. correct??

I essentially just plotted the above points onto a graph, and using the formula squareroot((10^2)+(8^2)) came up with the answer 13 km/h. And then using the tangent function = 8/10 = 39 degrees.

Should I be doing something else, or is the method I am using correct?

Thanks

 

[learningphysics]

Question: Boat traveling south at 10 km/h. A seaman on deck measure wind velocity to be 8 km/h east. What is the actual wind velocity? (the velocity of the wind relative to the ground)

Basically your statement $$\overrightarrow{v_{wb}}$$ = $$\overrightarrow{v_{w}}$$ - $$\overrightarrow{v_b}$$ would be the same as saying True Wind Speed = Apparant Wind Speed + Boat Wind Speed. correct??

You meant boat speed, not boat wind speed right? We should use velocity (since velocity is the vector and speed is the scalar)...

True Wind Velocity = Apparant Wind Velocity + Boat Velocity. Yes, that's right.

I essentially just plotted the above points onto a graph, and using the formula squareroot((10^2)+(8^2)) came up with the answer 13 km/h. And then using the tangent function = 8/10 = 39 degrees.

Should I be doing something else, or is the method I am using correct?

Yes, your method is right. You got the right magnitude and direction, so I think everything is correct. Hope I didn't confuse you too much.

 

[Iniuria12]

Thanks a bunch for your help\insight. That is definatly a weight off of my shoulders. For the past few hours I had the correct answer the entire time, and I've been teaching myself physics and haven't had anyone to verify what I've done. Appreciate your time and help once again.

Cheers.

 

[learningphysics]

Thanks a bunch for your help\insight. That is definatly a weight off of my shoulders. For the past few hours I had the correct answer the entire time, and I've been teaching myself physics and haven't had anyone to verify what I've done. Appreciate your time and help once again.

Cheers.

You're welcome.