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

AI Thread Summary
The discussion centers on solving for the position vector using 2-D vector analysis, specifically addressing confusion around the velocity vector's role in the calculations. Participants clarify that the velocity vector is derived from the graph, which indicates direction, and that the j-component of the position vector does not necessarily need to be zero for the angle θ to be zero. The original poster initially misunderstands the relationship between the position vector and the angle of the velocity vector, leading to confusion about the calculations. Ultimately, they realize their mistake was in misinterpreting the t vs. θ graph as a velocity graph. This highlights the importance of understanding the distinctions between different vector components in 2-D analysis.
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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