Relative Motion- Airplane vs radar station and tracking the displacement

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Homework Help Overview

The discussion revolves around a relative motion problem involving an airplane and a radar station, focusing on tracking the displacement of the airplane as it approaches the radar station. The problem includes specific distances and angles related to the airplane's position at two different times.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the vector representation of the airplane's position, with one suggesting a head-to-tail method for vector addition. Others raise concerns about inconsistencies in the problem's parameters, such as angles and distances, and propose using the law of cosines for calculations.

Discussion Status

The conversation includes attempts to clarify the problem setup and calculations, with some participants questioning the accuracy of the initial values used. There is an ongoing exploration of different methods to approach the problem, including vector subtraction and component analysis.

Contextual Notes

Participants note discrepancies in the given values for angles and distances, which may affect the calculations. The original poster acknowledges errors in their initial setup and provides corrected values for further discussion.

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Relative Motion-- Airplane vs radar station and tracking the displacement

In Figure 4-49, a radar station detects an airplane approaching directly from the east. At first observation, the airplane is at distance d1 = 340 m from the station and at angle θ1 = 34° above the horizon. The airplane is tracked through an angular change Δθ = 122° in the vertical east–west plane; its distance is then d2 = 800 m. Find the (a) magnitude and (b) direction of the airplane's displacement during this period. Give the direction as an angle relative to due west, with a positive angle being above the horizon and a negative angle being below the horizon.
http://edugen.wiley.com/edugen/courses/crs4957/art/qb/qu/c04/fig04_55.gif
2. The attempt at a solution

I set the vectors head to tail, where a is the vector to the original sighting, b is the vector to the second sighting. So I did this:
[PLAIN]http://i629.photobucket.com/albums/uu15/amorxamor/blahhcopy.jpg I really really really appreciate any help :c
 
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Some odd things in the solution:
- the angle of the initial vector to the plane is given as 30 degrees but you have it marked as 34 degrees on the diagram and in the trig functions.
- the initial distance to the plane is given as 360m but you use 340 in the calcs
- d2 is given as 780, but you use 800 in the calcs

The x,y component approach should work.
Maybe easier to use the law of cosines on the triangle to answer part (a) in one step.
 


Delphi51 said:
Some odd things in the solution:
- the angle of the initial vector to the plane is given as 30 degrees but you have it marked as 34 degrees on the diagram and in the trig functions.
- the initial distance to the plane is given as 360m but you use 340 in the calcs
- d2 is given as 780, but you use 800 in the calcs

The x,y component approach should work.
Maybe easier to use the law of cosines on the triangle to answer part (a) in one step.

yeah, I got the problem wrong and it gave me new values sorry I can't believe I didn't catch that...

d1 (|a|) should be 340 m

theta 1 should be 34 degrees

delta theta should be 122 degrees

d2 (|b|)= 800m
 


I disagree in the last part. I think you need to subtract the two vectors rather than adding them:
-730.8i +325.4j - (281.9i + 190.1j)
The magnitude comes out to 1022 and it checks by law of cosines.
 

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