Position and Displacement Vectors- The Watermelon Seed Vector

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Homework Help Overview

The discussion revolves around calculating the position and displacement vectors of a watermelon seed based on its coordinates in a three-dimensional space. The original poster presents specific coordinates and seeks to determine the magnitude and angle of the position vector, as well as the displacement vector after a movement to new coordinates.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for the position vector's magnitude and angle, with attempts to clarify the displacement vector's angle. There are questions about the correct approach to finding the angle of displacement, including the use of inverse tangent and sine functions.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning the methods used to find the angle of the displacement vector. Some guidance is offered regarding visualizing the vectors in a coordinate system, but there is no consensus on the correct approach for part (d).

Contextual Notes

There is mention of difficulties in visualizing the displacement vector and a note that the z-coordinate is not relevant since it remains constant at zero for both positions. The original poster expresses frustration with the calculations not being accepted by the homework platform.

Pmand92
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A watermelon seed has the following coordinates: x = -6.6 m, y = 2.2 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 (8.2 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?


A) Magnitude was found using (-6.6)^2+(2.2)^2=sqrt(48.4)=6.96m or 7.0m

B) The angle was found using inverse tangent of (2.2/6.6)=18.4 degrees. I subtracted 18.4 from 180 to get 161.6.

C) The magnitude was found using the displacement equation (Δx=X2-X1). 8.2m-(-6.6m)=14.8m

Up until part D, all of the answers are correct in WileyPlus. I have tried multiple ways trying to find this angle however, no success. Perhaps I am looking past the basic point. Please help.
 
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For (c), there is a change in y position too.
 
I know, it's -2.2m. I forgot to mention it.
 
Hi, Pmand92. Can you show us the details of what you did for part (d)? What did you get for an answer?
 
Well I tried using the magnitude (14.8 squared) minus 2.2 squared to find a component so I could try the inverse tangent to find the angle. tan^-1(2.2/14.8)=8.5 degrees and then I subtract that from 180, which 171.5. Then at this point, I tried to put it into WileyPlus, which was wrong. So I tried inverse sin with just 2.2/14.8, which didn't work. At this point, I have switched out numbers just trying to see if I could get an answer for the WileyPlus to take the answer.
 
To be honest, I can't remember all of the ways to find the angle.
 
The displacement is a vector pointing from the initial position to the final position. Draw a sketch to see how that vector is oriented relative to the positive x direction.
 
I am having a hard time visualizing that.
 
Draw an x-y cartesian coordinate system and locate the initial and final locations (positions) of the seed. The displacement will be a vector pointing from the initial location to the final location. You don't need to worry about the z-coordinate since it's 0 for both the intial and final postitions.
 

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