Physics Help Forum Airplane relative velocity

 Feb 10th 2013, 07:36 PM #1 Member   Join Date: Jan 2013 Posts: 42 Airplane relative velocity An airplane pilot sets a compass course due west and maintains an airspeed of 220 km/h. After flying for a time of 0.500 h, she finds herself over a town a distance 120 km west and a distance 20 km south of her starting point Find the wind velocity vp/e= velocity of plane relative to earth vp/a = velocity of plane relative to air va/e= velocity of air relative to earth Vp/e= vp/a +va/e This is all I have and I am stuck. Vp/e=220km/h+va/e I also did the pythagorean theorem with the distances given to get my resultant. R=sqrt[ (120km)^(2) + (20km)^(2) ] R=122 km. Not sure what to do with this though. Could somebody point me in the right direction?
 Feb 11th 2013, 06:35 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 You need to remember that this requires vector addition (and subtraction). Starting with Vp/e = Vp/a + Va/e, rearrange to get: Va/e = Vp/e - Vp/a. The value for Vp/a is 220 Km/hr due west, or in cartesian coordinates: (-220i + 0j) Km/hr, where 'i' is the unit vector in the +x direction (+i = east and -i = west) and 'j' is the unit vector heading north (and hence -j is heading south). The vector Vp/e is found by determining the amount of ground covered per hour, which is (-240i - 40j)Km/hr; this comes from the position of the plane after 1/2 hour of travel. So we have: Va/e = Vp/e - Vp/a = (-240i - 40j)Km/hr - (-220i + 0j)Km/hr = (-20i -40j)Km/hr The magnitude of this wind velocity can be found using Pythagoras. Last edited by ChipB; Feb 11th 2013 at 06:43 AM.
Feb 11th 2013, 01:56 PM   #3
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 Originally Posted by ChipB You need to remember that this requires vector addition (and subtraction). Starting with Vp/e = Vp/a + Va/e, rearrange to get: Va/e = Vp/e - Vp/a. The value for Vp/a is 220 Km/hr due west, or in cartesian coordinates: (-220i + 0j) Km/hr, where 'i' is the unit vector in the +x direction (+i = east and -i = west) and 'j' is the unit vector heading north (and hence -j is heading south). The vector Vp/e is found by determining the amount of ground covered per hour, which is (-240i - 40j)Km/hr; this comes from the position of the plane after 1/2 hour of travel. So we have: Va/e = Vp/e - Vp/a = (-240i - 40j)Km/hr - (-220i + 0j)Km/hr = (-20i -40j)Km/hr The magnitude of this wind velocity can be found using Pythagoras.
Thank you!

Feb 12th 2013, 02:10 PM   #4
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Join Date: Jan 2013
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 Originally Posted by ChipB You need to remember that this requires vector addition (and subtraction). Starting with Vp/e = Vp/a + Va/e, rearrange to get: Va/e = Vp/e - Vp/a. The value for Vp/a is 220 Km/hr due west, or in cartesian coordinates: (-220i + 0j) Km/hr, where 'i' is the unit vector in the +x direction (+i = east and -i = west) and 'j' is the unit vector heading north (and hence -j is heading south). The vector Vp/e is found by determining the amount of ground covered per hour, which is (-240i - 40j)Km/hr; this comes from the position of the plane after 1/2 hour of travel. So we have: Va/e = Vp/e - Vp/a = (-240i - 40j)Km/hr - (-220i + 0j)Km/hr = (-20i -40j)Km/hr The magnitude of this wind velocity can be found using Pythagoras.
How can you tell when a velocity is relative to the ground? Is it because they gave you distance and time? I am having trouble determining whether something is relative to the air or ground when the problem does not exactly give it to you.

 Feb 12th 2013, 02:21 PM #5 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 The question said in part: "she finds herself over a town a distance 120 km west and a distance 20 km south of her starting point." It seems to be pretty clear that the distance discussed here is relative to the ground, as towns typically don't shift with the wind. Of course there may be exceptions, but in general when trying to understand homework problems it's best to keep with the simplest concept.
Feb 12th 2013, 02:35 PM   #6
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 Originally Posted by ChipB The question said in part: "she finds herself over a town a distance 120 km west and a distance 20 km south of her starting point." It seems to be pretty clear that the distance discussed here is relative to the ground, as towns typically don't shift with the wind. Of course there may be exceptions, but in general when trying to understand homework problems it's best to keep with the simplest concept.
I see and thanks again.

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