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# Can anyone solve the following problem? (Vectors)

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A plane is heading due north: its nose points towards the north direction, but its trajectory on the ground deviates from the north direction due to a sideways component of the wind. The plane is also climbing at a rate of 120 km/h (height increase per unit time). If the plane's airspeed is 490 km/h and there is a wind blowing 120 km/h to the northwest, what is the ground speed of the plane?

Apr 5, 2019

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Josh, this is a simple question.

Would you know how to solve this if the aircraft wasn’t ascending?

The solution method uses the law of cosines

$$\Large c^2 = a^2 + b^2 - 2ab \cos \normalsize (C)$$

My cat, who learned this from my dog, can solve it while catnapping.

Present your attempted solution (in coherent form) and my cat will respond with a critique.

GA

Apr 5, 2019
edited by GingerAle  Apr 5, 2019
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https://artofproblemsolving.com/wiki/index.php/Law_of_Cosines

Guest Apr 5, 2019
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You know Mr. BB, if you want to be a flasher, you should wear a raincoat ...and carry a magnifying glass.

GA

GingerAle  Apr 5, 2019
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Here is what MY dog finds as the answer (he may not be as bright as GA's dog...so he may be incorrect)

The WIND speed does not affect the AIRspeed...it DOES affect the GROUNDspeed

Vertical speed (z component) is also not affected by the wind speed and is equal to ( given )  120  (km/hr)

The y component of the airspeed is found with pythagorean theorem     490^2 = 120^2 + y^2     y = 475 km/hr

The x component of the ground speed is .707 (120) = 84.84    (this component is due to the wind ONLY)

The y component of the wind speed is     .707 (120) = 84.84   and is additive to the 475 km/hr

Only the x and y components contribute to the ground speed....the vertical speed does not

the x and y components  squared and added together are the resultant ground speed squared

x^2 + y^2  = GS^2

84.84^2  + (84.84 + 475)^2       = GS^2                  (yah?...no guarantee!)

GS = 566.2 km/hr

Apr 5, 2019
edited by ElectricPavlov  Apr 5, 2019
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DC Thomas Copper waking from his catnap and critiquing your post.  Translation from Purrrsian:  W T F ...

This plane is flying into a headwind, and it is climbing like a bat leaving heII.

This means the ground speed is less than the air speed.

GA

GingerAle  Apr 5, 2019
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My dog just pointed out to my cat that the wind direction is presented (unconventionally) as a bearing, and not as a source, making it a partial tail wind.  EP’s solution would be correct if the plane was not ascending.

GA

GingerAle  Apr 6, 2019
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GA.....     the ascending speed has nothing to do with the ground speed....it is only used to find the component of the AIRspeed (490 km/hr) that is in the 'y' (north direction) ....

Which is added to the y component of the WINDspeed (1)

The x component of the windspeed   is then used with  (1)  to find the GROUNDspeed......

That's a crazy cat!     ~EP

ElectricPavlov  Apr 6, 2019
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Sorry EP. I just did the calculation and I can see that you have the same result as I do.

$$GS = \sqrt{TAS^2 - {V_C}^2 + {V_W}^2 + 2 \cdot \sqrt{TAS^2 - {V_C}^2}\cdot V_W \cdot cos(\Phi)}\\ \\ \text{ }\\ \sqrt{470^2 - 120^2 + 120^2 + 2 \cdot \sqrt{470^2 - 120^2}\cdot 120 \cdot cos(315)} \approx 566.3\\$$

Site calculator input:

a=475; v=120; w=120; t=315; sqrt(a^2 – (v)^2 + (w)^2 + 2 *\sqrt(a^2 - v^2)*w * cos(t))

I only glanced at your work and when I saw this:

...the ascending speed has nothing to do with the ground speed (This is true, but it does affect airspeed, and airspeed is used to calculate ground speed.)

and I just assumed you’d not corrected for the vertical component.

I’ll be drop-kicking my cat into next week for this screw-up. I need to figure out the proper temporal launching vector, else a week later he’ll return to normal time, landing on my head with claws fully extended for braking. My dog thinks it’s quite funny!

Do you have any ideas for a temporal launching vector? There may not be one ... this just might be temporal karma.  GA

GingerAle  Apr 7, 2019