Calculating Airplane Velocity and Direction with Wind

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To calculate the correct heading for an aircraft flying due west in a 50 km/h south wind, the pilot must consider both the aircraft's airspeed of 205 km/h and the wind's velocity. The initial calculation suggested a heading of 13.7 degrees north of west, but this approach is incorrect as it doesn't account for vector addition. The proper method involves determining the direction of the aircraft's velocity vector so that the resultant vector points directly west. A diagram can help visualize the vectors, and using trigonometry will yield the correct angle for the aircraft's heading. The final speed relative to the ground should also be recalculated based on the correct vector direction.
Canesfanatic
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Hi, this is my first time posting here, and i was working out this physics problem and it occurred to me that when i calculated the velocity of an airplane in my problem, it's higher than the speed of the airplane...i know i probably didn't explain that too well, so here goes:

the problem:
The pilot of an aircraft wishes to fly due west in 50.0 km/h wind blowing toward the south. The speed of the aircraft in the absence of a wind is 205 km/hr.
a. In what direction should the aircraft head?
on this, I set up a right triangle with the velocities as vectors and calculated that the aircraft should fly 13.7 degrees north of west using angle=tan^-1 * (50.0 km/h/205km/h

b. What should its speed be relative to the ground?
Again, on this I set up another right triangle, and when i calculated the answer, it came up as 211. km/hr. I used the pythagorean theorum with (50.0km/h)^2+(205km/h)^2= C^2

What I am not sure about is whether I went about this problem with the right approach and whether 211. km/hr is the correct answer for B. All help is appreciated :biggrin:
 
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Canesfanatic said:
a. In what direction should the aircraft head?
on this, I set up a right triangle with the velocities as vectors and calculated that the aircraft should fly 13.7 degrees north of west using angle=tan^-1 * (50.0 km/h/205km/h
This isn't quite right. You need to add these vectors: (1) The velocity of the plane with respect to the air + (2) the velocity of the air with respect to the ground. This gives you the resultant velocity of the plane with respect to the ground--which you know must point west. So you need to find what direction (1) must point so that the resultant is purely to the west. You know the magnitude of (1), now find the direction.

Draw yourself a diagram: Start by drawing vector (2). Then play around with vector (1) until you see where it must point--then use a little trig to find the angle.
 
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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