Relative Velocities: Plane Displacement from City in 2h Time

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SUMMARY

The discussion focuses on calculating the displacement of a plane flying above a city, factoring in wind velocity. The wind blows at 72 km/h south, while the plane travels at 320 km/h at an angle of 28 degrees south of west relative to the wind. The resultant velocity of the plane relative to the Earth is determined to be 248 km/h. The participants emphasize the importance of vector addition in determining the new angle of the plane to the Earth, suggesting the use of the sine law for this calculation.

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Homework Statement


The wind is blowing at 72km/h south relative to the earth, and a plane is traveling 320km/h (28 degrees south of west relative to the wind) directly above a city. give displacement from the city in 2h time


Homework Equations


Plane is P wind is W Earth is E

VPW=VPE+VWE



The Attempt at a Solution



re arrange to find that
VPE=VPW-VWE
VPE=248km/h

The only thing I don't understand is how to determine the new angle of the plane to the earth...
 
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Velocities are vectors. You cannot add or subtract them directly.
 
Hi slpnsldr. Your formula was incorrect. It should look like this
\vec{_{p}V_{E}}=\vec{_{p}V_{w}}+\vec{_{w}V_{E}}

Notice when you add these relative velocities, the subscript "w" cancel each other out. And that is the only trick to it.
 
ok, I've figured out my displacement.. but I am still unclear about how to find the planes new angle
somethign to do with the sine law?
 
slpnsldr said:
ok, I've figured out my displacement.. but I am still unclear about how to find the planes new angle
somethign to do with the sine law?

Oh, that might be hard to explain without a diagram. Maybe you need to include a diagram.
 

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