How do I solve for the position vector using 2-D vector analysis?

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Homework Help Overview

The discussion revolves around solving for the position vector using 2-D vector analysis, specifically examining the relationship between the velocity vector and the position vector over time. Participants are analyzing a problem presented in a graphical format.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the reasoning behind using the velocity vector, questioning its relevance to the position vector. They explore the implications of the j-component of the position vector and its relationship to the angle θ at a specific time.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts and clarifying misunderstandings. Some have provided insights into the relationship between the velocity and position vectors, while others express confusion about the implications of the graphs presented.

Contextual Notes

There is mention of a graph that illustrates the velocity vector and its angle, which is central to the participants' reasoning. The original poster indicates a lack of clarity regarding the relationship between the components of the position vector and the angle θ.

princeton_wu
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Homework Statement



http://imgur.com/wNusHOw

Homework Equations



I have the solutions and how they did it. THey took the deriv for the velocity vector, and then using t=0 and t=14, they found e=3.5 and f=-0.125

The Attempt at a Solution



I understand the math, but I don't understand why this is correct
1) why did they take the velocity vector?
2) if I plug e and f back into the position vector, and using t=14 i should get θ=0, ie, j-component is 0. But I don't. So what am I doing wrong?
 
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welcome to pf! :smile:

hi princeton_wu! welcome to pf! :smile:
princeton_wu said:
1) why did they take the velocity vector?

because the graph gives you the direction of the velocity vector
2) if I plug e and f back into the position vector, and using t=14 i should get θ=0, ie, j-component is 0. But I don't.

yes you do …

j-component = e + 2ft = 3.5 - 3.5 = 0 :wink:
 
tiny-tim said:
hi princeton_wu! welcome to pf! :smile:because the graph gives you the direction of the velocity vectoryes you do …

j-component = e + 2ft = 3.5 - 3.5 = 0 :wink:

i mean the r vector; shouldn't the J-component of the r-vector be 0 @ t=14? this way, the angle @ t=14 would be 0.
 
princeton_wu said:
i mean the r vector; shouldn't the J-component of the r-vector be 0 @ t=14? this way, the angle @ t=14 would be 0.

i'm not following your reasoning :confused:

the graph shows that vj = 0 (because θ = 0) at t = 14, it says nothing about r :smile:
 
tiny-tim said:
i'm not following your reasoning :confused:

the graph shows that vj = 0 (because θ = 0) at t = 14, it says nothing about r :smile:

sorry, I'm confused too :-p

If you use t=14 and plug it in the original position vector, shouldn't the J-component of the position vector be 0? Reasoning being that for θ to be 0, the J-component has to be 0?

thanks for your patience tiny tim!
 
(just got up :zzz:)
princeton_wu said:
Reasoning being that for θ to be 0, the J-component has to be 0?

but θ (given in the graph) is stated to be the angle of the velocity vector …

i don't understand what you think that has to do with the position vector :redface:
 
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I mulled over it last night and I finallly got it. My problem stemmed from the fact that I didn't realize that a t vs θ graph is a velocity graph). Thanks Tim! :smile:
 

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