Position Vector & Displacement: Particle Homework

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Homework Help Overview

The problem involves determining the position vector and displacement of a watermelon seed based on its coordinates in a three-dimensional space. The original poster seeks to find both the magnitude and angle of the position vector relative to the positive x-axis, as well as the displacement after moving to a new set of coordinates.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the components of the displacement vector and the corresponding angle. There are questions about how to derive the angle without having one of the angles initially, and whether the angle should be adjusted based on its relation to the positive x-axis.

Discussion Status

Participants have provided various approaches to finding the angle of the displacement vector, including using inverse tangent and considering the orientation of the angle relative to the positive x-axis. There is recognition of potential issues with automated grading systems, and some participants suggest discussing discrepancies with instructors.

Contextual Notes

There is mention of a specific grading system (Wiley Plus) that may not accept certain angle representations, which adds complexity to the discussion. Participants are navigating the constraints of the problem while trying to clarify their understanding of vector components and angles.

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Homework Statement


A watermelon seed has the following coordinates: x = -8.1 m, y = 8.8 m, and z = 0 m. Find its position vector as (a) a magnitude and (b) an angle relative to the positive direction of the x axis. If the seed is moved to the xyz coordinates (3.3 m, 0 m, 0 m), what is its displacement as (c) a magnitude and (d) an angle relative to the positive direction of the x axis?


Homework Equations





The Attempt at a Solution


a= 11.9 Meters
b= 133 degrees
c= 15.2 Meters
d = ?

How do I find this final angle??
 
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Find the x and y components of the displacement vector and get the angle the usual way.
 
All I have is the magnitude of the displacement vector. I know I need to find the x and y components of it, but how do I do that without at least having one of the angles? Or do i already have it??
 
You have the coordinates of two points. The displacement is a vector whose components is the difference finish - start, (x2-x1, y2-y1, z2-z1)
 
so my displacement vector's components are (11.4, -8.8, 0)? So the angle would be inverse tan of (-8.8/11.4)?
 
Correct.
 
the angle I get from that calculation is -37.67..Now it says it wants an angle relative to the positive direction of the x-axis. Would that be it, or do I have to subtract that number from something else?
 
Draw the point on a sheet of paper. Connect it with a straight line to the origin. Then look at the angle that starts on the positive x-axis and goes counterclockwise to the line that you have drawn. That's the one you want. What is its relation to 37.67 degrees?
 
ok, is it 360-37.67? which is going to be 322.33
 
  • #10
I put in 322.33, and it came out wrong! The correct angle was -37.67..Bummer
 
  • #11
Bummer indeed. If this is one of those evil Web-based, calculated type questions like Webassign or Mastering Physics, I think you should explain your answer to your instructor. Chances are he/she will override the machine. I know I would. :wink:
 
  • #12
yea wiley plus..Horrible system!
 
  • #13
thanks anyways for your help!
 

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