Relative Plane Motion

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Homework Statement


You are traveling on an airplane. The velocity of the plane with respect to the air is 110 m/s due east. The velocity of the air with respect to the ground is 39 m/s at an angle of 30° west of due north.

What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due east).

Homework Equations


None.

The Attempt at a Solution


Ok, so I already obtained the speed by finding ##\vec V_{ag} = -39\cos(30^\circ)\hat i + 39\sin(30^\circ)\hat j##, and then adding it to ##\vec V_{pa} = 110\frac{m}{s}\hat i##, yielding ##V_{pg} = 95.73\frac{m}{s}##. The component form is ##\langle 76.23 , 19.5 \rangle##.

I then evaluated ##\arctan(\frac{19.5}{76.23})##, giving me ##\theta = 14.35^\circ##.

I don't understand where to go from here. What do they mean by letting "##0^\circ## represent due north" and "##90^\circ## represent due east"?

Thanks.
 

Answers and Replies

  • #2
andrewkirk
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What do they mean by letting "0∘0^\circ represent due north" and "90∘90^\circ represent due east"?
The international convention for air navigation is that the heading of a plane is indicated by an angle measured as a clockwise rotation from North. Note that that is the opposite direction from the (anti-clockwise) rotation convention employed in the number plane, and it also starts from a different point (North/'Up' rather than 'Right'). What they mean is that they want you to use that navigation convention to give your answer. So if your answer was South-East (it won't be) it should be stated as '135 degrees'.

Your calculation of ##\vec{V}_{ag}## is wrong. If the heading is 30 degrees West of North then the direction vector is in the second quadrant and makes an angle of 30 degrees with the vertical axis.
 
  • #3
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Your calculation of ##\vec{V}_{ag}## is wrong. If the heading is 30 degrees West of North then the direction vector is in the second quadrant and makes an angle of 30 degrees with the vertical axis.
Ah, you're right! It should be ##\vec V_{ag} = 39\cos(30^\circ)\hat i + 39\sin(30^\circ)\hat j##, right?
 
  • #4
haruspex
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Your calculation of ##\vec{V}_{ag}## is wrong. If the heading is 30 degrees West of North then the direction vector is in the second quadrant and makes an angle of 30 degrees with the vertical axis.
Yes and no. The solver is free to map x and y co-ordinates to compass directions in any consistent manner. The OP's expression is right if S is mapped to the positive X axis and W to the positive Y axis. But you are probably right that this was simply an error in this case.
 
  • #5
haruspex
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Ah, you're right! It should be ##\vec V_{ag} = 39\cos(30^\circ)\hat i + 39\sin(30^\circ)\hat j##, right?
Not if you are mapping N to +Y and E to +X, i.e. the obvious mapping.
 
  • #6
andrewkirk
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The OP's expression is right if S is mapped to the positive X axis and W to the positive Y axis.
I think that's ruled out though, by the following assignment in the OP for the plane's Due East heading:
##\vec{V}_{pa}=110 \frac{m}{s}\hat{i}##
 
  • #7
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Alright, so apparently the equation I'm supposed to use is ##V_{pg} = V_{pa} + V_{ag}##. However, now I get the incorrect answer for the speed. The calculation I initially obtained (95.73 m/s) was correct.

##V_{pg_x} = V_{pa_x} + V_{ag_x} = 110 \frac{m}{s} + 39\cos(30^\circ) = 143.77\frac{m}{s}##

##V_{pg_y} = V_{pa_y} + V_{ag_y} = 0 + 39\sin(30^\circ) = 19.5\frac{m}{s}##

So I get the same results for each component. However, I have to use the Pythagorean theorem at this point (according to my professor). Which makes sense, considering that they are asking for the speed (magnitude) rather than the velocity.

##\sqrt(143.77^2 + 19.5^2) = 145.09\frac{m}{s}##

According to SmartPhysics, this is not the correct speed.

I am incredibly confused now.
 
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Alright, I see what you guys were saying. The angles are supposed to be 120 degrees; however, I still get the wrong result (96.6).
 
  • #9
haruspex
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Alright, I see what you guys were saying. The angles are supposed to be 120 degrees; however, I still get the wrong result (96.6).
That's the right groundspeed, but the question asks only for the bearing, no?
 

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