Relative velocity problem. NEED HELP.

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The problem involves calculating the new velocity of a jet airliner moving at 300 mph due east in the presence of a wind blowing at 100 mph at an angle of 30° north of east. The solution requires vector addition, where the plane's velocity vector is combined with the wind's velocity vector to find the resultant velocity relative to the ground. By establishing a coordinate system with east as the positive x-direction and north as the positive y-direction, the components of each vector can be determined. The final equation for the resultant vector is R = A + B, where R is the new ground velocity. Collaborating with others can simplify the understanding of vector addition in this context.
sanghoon
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Homework Statement


A jet airliner moving initially at 300 mph due east enters a region where the wind is blowing at 100 miph in a direction 30° north of east. What is the new velocity of the aircraft relative to the ground?


Homework Equations


I don't know any equations for this question.


The Attempt at a Solution


I thought it might be like using the 300 mph... caculate the hypotenus velocity vector using the angle, and add 100 mph... I don't think it's right.
 
Last edited:
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This is a vector addition problem. One adds the planes velocity vector to the winds velocity vector. The plane is supported/carried by the wind and the wind is moving, which is much like a boat in a river with a current.

So pick a coordinate system, and the determine the components of each vector.

Pick the direction to East as +x, and the direction to North as +y.
 
Oh wait... so if the plane's velocity vector was A, and the wind's velocity vector was B.
It would be B + A = R?
R being the new velocity vector relative to the ground.
 
Last edited:
sanghoon said:
Oh wait... so if the plane's velocity vector was A, and the wind's velocity vector was B.
It would be B + A = R?
R being the new velocity vector relative to the ground.
Correct!
 
Thank you very much. Much easier when you have another to think with.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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