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2D Vectors  Addition and Subtraction of Successive Displacement Vectors 
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#1
Jan2413, 05:28 PM

P: 4

The figure shows the successive displacements of an aircraft flying a search pattern. The initial position of the aircraft is P and the final position is P'. What is the net displacement (magnitude and direction) between P and P'?
My attempt: So, my understanding is that the net displacement of the successive displacement vectors is the overall sum of the successive displacement vectors. (?) Also, referring to the rules of alternate interior angles, ∠A, ∠B, and ∠C = 60° (?) ΔA Hypotenuse = 18km Height = 18sin60° Base = 18cos60° ΔB Hypotenuse = 9.5km Height = 9.5sin60° Base = 9.5cos60° ΔC Hypotenuse = 12km Height = 12sin60° Base = 12cos60° My attempt to find the base of the triangle formed by the point P and P' and the xaxis was to add the bases of ΔA and ΔB and subtract the base of ΔC from the sum. With my calculations (using rounded decimals and plugging in the exact values): I got ~13km for →P (incorrect because my book reads 11.2km) and for ∠θ I got 57.1° (incorrect, because the remaining angle should be 27.7° and is 32.9° with my calculations) I am doing something wrong. HELP 


#2
Jan2413, 11:46 PM

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P: 6,653

AM 


#3
Jan2513, 12:07 AM

P: 616

In this particular case, you're using the wrong angle (assuming 60° is the angle between the [itex]y[/itex]axis and the first vector). 


#4
Jan2513, 12:10 AM

P: 616

2D Vectors  Addition and Subtraction of Successive Displacement Vectors



#5
Jan2513, 09:16 AM

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P: 6,653

Part of the problem here is that the diagram is misleading. The vector lengths and angles do not fit. For example, the heights of ΔA and ΔB are the same. Using the figures given, however, the height of ΔA is 18sin(30) =9 km and the height of ΔB 9.5cos(30) = 8.22 km. (In other words, arctan 9.5/18 = 27.8 deg ≠ 30 deg.). I would suggest that you add the vector components using the angles given and disregard the triangles, as tms has suggested. The answer given is correct (11.2). AM 


#6
Jan2513, 09:41 AM

P: 616




#7
Jan2513, 11:35 AM

P: 4

I will upload the original diagram from my book. Maybe I will be able to better visualize it rather than using my quick sketch of the diagram drawn in MS paint. Thanks. 


#8
Jan2513, 12:52 PM

P: 616




#9
Jan2513, 02:24 PM

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P: 6,653

AM 


#10
Oct3113, 06:46 PM

P: 8

I had to solve the same problem. I finally solved it but not completely. I am not able to determine the angle that PP' makes.
I would appreciate some help. I am not able to figure this part out. Thanks. 


#11
Oct3113, 06:50 PM

P: 8

Please note that in the figure in post #1, all the 60 degree angles are supposed to 30 degrees.



#12
Oct3113, 07:15 PM

P: 616




#13
Oct3113, 08:02 PM

P: 8

I had to redraw the figure and this took me a while. Alright, here is my work.
Angle APz is 60 degrees and AP is 18km. Py = 18 cos 60 = 9km. This makes Ay = 15.6km. Angle BAx is 60 degrees and AB is 9.5km. Ax = (cos 60) 9.5 = 4.75 km. This makes Bx = 8.22km. xy = AyAx = 10.85km. Angle wBP' is 60 degree and BP' is 12 km. wP' = (sin 60) 12 = 10.39km. wB = (cos 60) 12 = 6km. The x  coordinate for B is Py+Bx = 9+8.22 = 17.22km. This makes Pz 17.22wB = 17.226 = 11.2km. zP' = wP'wz = wP'xy = 10.3910.85 =  0.41km This means that P' coordinates are 11.2km, 0.41km. This makes PP' square root of (11.2)^2  (0.41)^2 = square root of 125 = 11.2km. 


#14
Oct3113, 08:06 PM

P: 8

My final answer is PP' is 11.2 km and 2.1 degrees North of East.



#15
Oct3113, 09:58 PM

P: 616

A couple of pieces of advice: First, don't use x, y, and z to label point; those letters are too easily confused with the coordinates. For instance, your Ay is really the x coordinate of point A.
Second, in this problem, you are given three vectors in polar coordinates (length and direction), which you must add together. The easiest way is to convert each vector to rectangular coordinates (x, y) and then simply add them. This will eliminate all the extra points you created, and if you use the right angles, will take care of the signs automaticaly. [tex] \begin{align} BP' &= (r, \vartheta) = (12, 210^\circ) \\ &= (r \cos \vartheta, r \sin \vartheta) \\ &= (10.4, 6) \end{align}[/tex] 


#16
Oct3113, 10:19 PM

P: 8

What would the magnitude of P' be?



#17
Oct3113, 10:34 PM

P: 8

The coordinates of A is 9,15.6.
B is 17.2, 10.85 and P' is 11.2 and 0.41. The object first moves to A then B and then P'. So, do I have to subtract AB? If I do that the coordinates of C and the final answer is very different. 


#18
Oct3113, 10:45 PM

P: 616

For instance, the vector from A to B has a length of 9.5 and a direction of 30 degrees (in polar coordinates), which leads to x and y components of (8.22, 4.75). Do that for all three vectors, and then add them. 


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