IBV10 using displacement vectors

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    Displacement Vectors
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Discussion Overview

The discussion revolves around the use of displacement vectors in a problem involving motion, specifically focusing on deriving a Cartesian equation for a path. Participants explore the calculation of distance and velocity, as well as the relationship between points and slopes in the context of linear motion.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant calculates the distance using the Pythagorean theorem, arriving at a distance of 20 km but expresses uncertainty about how a velocity of 13 km/min was derived.
  • Another participant suggests that the velocity can be computed by taking the derivative of the position vector, indicating a need for calculus in the analysis.
  • There is a proposal that if the position function is linear, the velocity must be constant, although this is met with skepticism by the same participant.
  • A participant asserts that if the velocity is constant, they can derive a slope from given points, leading to the equation $5x + 12y = 224$ based on their conjecture.
  • Another participant requests clarification on the other point needed to find the line between two points, indicating a need for more information to fully address the problem.
  • Participants discuss the derivation of the Cartesian equation from a given point and slope, reiterating the equation $5x + 12y = 224$ as a result of their calculations.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of calculus for the problem and the assumptions regarding linear motion. There is no consensus on the derivation process or the correctness of the proposed solutions.

Contextual Notes

Some assumptions about the linearity of the position function and the nature of the velocity remain unverified. The discussion also reflects uncertainty about the completeness of the problem statement and the necessary points for deriving the equation.

karush
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https://www.physicsforums.com/attachments/1341

for (i) just did $$16^2+12^2=400$$ so $$\sqrt{400} = 20$$ km

but don't know how they got the $$13 \frac{km}{min}$$

there are some more ? on this involving the displacement vectors but want to get this (a) part done first.
 
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karush said:
https://www.physicsforums.com/attachments/1341

for (i) just did $$16^2+12^2=400$$ so $$\sqrt{400} = 20$$ km

but don't know how they got the $$13 \frac{km}{min}$$

there are some more ? on this involving the displacement vectors but want to get this (a) part done first.
I can't think of a way to do this without Calculus. Well, here it is anyway.

[math]v = \frac{dr}{dt}[/math]

So take the derivative of r, compute v, then find the "size" of v.

-Dan
 
I suppose one could assume that given a linear position function, the velocity must be constant...:D
 
MarkFL said:
I suppose one could assume that given a linear position function, the velocity must be constant...:D
Nah. Too simple.

Good catch.

-Dan
 
OK, if the velocity is constant.
then there is a slope, since $$t\pmatrix{12 \\ -5}$$ is Time x Rate then$$ \sqrt{(12)^2+(-5)^2}= 13$$ km/min

now we have a point $(16, 12)$ and $m=-\frac{5}{12}$
from which we can derive the eq of
$5x+12y=224$

if correct, I did this by conjecture, not knowing the formal process.

still some more ? on this...:cool:
 
karush said:
now we have a point $(16, 12)$ and $m=-\frac{5}{12}$
from which we can derive the eq of
$5x+12y=224$
In the future, please post all of the question you need help with.

You seem to be trying to find the line between two points, one of which is (16, 12). What is the other point?

-Dan
 
topsquark said:
In the future, please post all of the question you need help with.

You seem to be trying to find the line between two points, one of which is (16, 12). What is the other point?

-Dan

ok, thot this was in the OP but it wasn't

The question was:
Show that the Cartesian equation of the path of Air One is:

$5x+12y=224$

However, from $r_1=\pmatrix{16 \\ 12} + t\pmatrix{12 \\ -5}$

we have point $(16, 12)$ and $m=-\frac{5}{12}$

we can derive $5x+12y=224$
 

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