How Is the Angle of Displacement Calculated in Vector Travel?

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The discussion focuses on calculating the angle of displacement for a plane's trip from city A to city C. The plane travels 400 km east and 940 km south, with the correct distances needing to be used rather than speeds. A right triangle is formed, allowing the use of trigonometry to find the angle, which is determined to be approximately 23 degrees east of south. Alternative angle representations include 67 degrees south of east or 157 degrees measured clockwise from north. The importance of accurately interpreting the problem's parameters is emphasized for correct calculations.
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A plane flies 400 km east from city A to city B in 45.0 min and then 940 km south from city B to city C in 1.50 h. (Assume i points east and j points north

In magnitude-angle notation, what is the plane's displacement for the total trip

I figured out the displacement is 1021.57 m but i can't figure out the angle. How is this done.
 
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I think what they mean by "magnitude-angle notation" is simply vector notation, so a magnitude and a direction must be given. Did you draw a picture? The first thing you do in a vector problem is draw a diagram, then it becomes easy. Use trigonometry to find the angle.
 
The answer to the question must be given in m and an angle, so it's not vector notation it's just two numbers straight out. I've tried using trig and keep getting the wrong answer.
 
show us the trig you've been doing, so we can see where it is wrong...
 
I've done arctan of 940/400, I've done arctan of 940/400 and tried adding and subtracting 180. It's all wrong.
 
Why are you using 400 and 940? Those are the distances in km the airplane would fly in one hour[\b]. The problem tells you that the airplane flies east at 400 km/h for only 45 minutes (what distance does it cover?) and flies south at 940 km/h for 1 1/2 hours (what distance does it cover?).
 
the 400 and 940 are only displacements, they aren't velocities. I thought you found the angle of displacement by using those numbers.
 
GingerBread27 said:
the 400 and 940 are only displacements, they aren't velocities. I thought you found the angle of displacement by using those numbers.

Ah, sorry, you are right. You don't really need the time then- I thought they were speeds.

Okay, If you draw a picture taking θ as the angle "east of south", you have a right triangle with "opposite side" has length 400 km and the "near side" has length 940 km. tan(θ)= 400/940= 0.4225 and so θ= 23 degrees.

Now! How are you writing the angle? As I said, that is 23 degrees "east of south". You could also give it as "67 degrees south of east" (90- 23= 67) or, measured clockwise from north as on a compass, as 157 degrees (180- 23= 157).
 
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