Relative Motion- Airplane vs radar station and tracking the displacement

[emmy]

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

 

[Delphi51]

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.

 

[emmy]

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

 

[Delphi51]

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.

 

[rwisz]

First of all, great job setting up the problem and drawing out the vectors! To find the magnitude and direction of the airplane's displacement, we can use the law of cosines:

c^2 = a^2 + b^2 - 2ab*cos(Δθ)

where c is the magnitude of the displacement. Plugging in the values from the problem, we get:

c^2 = (340)^2 + (800)^2 - 2(340)(800)*cos(122°)
c^2 = 115600 + 640000 - 435200*cos(122°)
c^2 = 755600 + 435200*cos(122°)
c^2 = 755600 + (-296734.72)
c^2 = 458865.28

Taking the square root, we get c = 677.18 m. So the magnitude of the airplane's displacement is 677.18 m.

To find the direction, we can use the law of sines:

sin(122°)/c = sin(θ2)/a

where θ2 is the angle of the displacement vector. Plugging in the values from the problem, we get:

sin(122°)/677.18 = sin(θ2)/340
sin(θ2) = (340*sin(122°))/677.18
sin(θ2) = 0.649
θ2 = sin^-1(0.649)
θ2 = 40.8°

Since we know that the displacement is in the east-west plane, we can say that the angle of the displacement vector is 40.8° relative to due west. To determine if it is above or below the horizon, we can use the fact that the angle of elevation of the airplane changes from 34° to 122°. Since 122° is greater than 90°, we know that the airplane must have moved above the horizon. Therefore, the direction of the displacement is 40.8° above the horizon.

In summary, the magnitude of the airplane's displacement is 677.18 m and the direction is 40.8° above the horizon and 40.8° relative to due west. I hope this helps!