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Old Feb 10th 2013, 07:36 PM   #1
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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?
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Old Feb 11th 2013, 06:35 AM   #2
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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.
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Old Feb 11th 2013, 01:56 PM   #3
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Originally Posted by ChipB View Post
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!
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Old Feb 12th 2013, 02:10 PM   #4
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Originally Posted by ChipB View Post
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.
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Old Feb 12th 2013, 02:21 PM   #5
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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.
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Old Feb 12th 2013, 02:35 PM   #6
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Originally Posted by ChipB View Post
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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