Velocity from position in a parametric

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Homework Help Overview

The discussion revolves around determining velocity from a position vector in a parametric context, specifically focusing on the relationship between position and velocity through derivatives and graphical representation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to graph position components to understand velocity, questioning the necessity of multiple graphs for different components. There is confusion regarding how to interpret the graphs to find velocity and when the particle is moving parallel to the axes.

Discussion Status

Participants are actively engaging with the problem, exploring graphical methods and discussing the implications of slopes in relation to velocity. Some guidance has been provided regarding the interpretation of graphs and the conditions under which the particle is stationary.

Contextual Notes

There is mention of a specific point in time where the tangent to the graph is parallel to the y-axis, as well as a hint regarding the condition for the particle being at rest, indicating the need for further exploration of these concepts.

-EquinoX-
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Homework Statement


http://img21.imageshack.us/img21/7910/68588225.th.jpg


Homework Equations





The Attempt at a Solution



I know that the velocity is the derivative of the position vector.. but I am kind of confused how to do this
 
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How about graphing the table of values to start?
 
e(ho0n3 said:
How about graphing the table of values to start?

I've done that as well... but from the graph I can't find the velocity
 
You should have drawn two graphs (why?) and the velocity is the slope (why?).
 
the velocity is the slope (why?).

because it's the derivative of the graph...

why two graph, I don't know
 
-EquinoX- said:
why two graph, I don't know
You need 2 because you have 2 position components: x vs t and y vs t.
 
ok, I got v = -4/5i + 5/5j, is this correct?

if that's correct, how do I answer this one:

Any time when the particle is moving parallel to the y-axis.
 
-EquinoX- said:
ok, I got v = -4/5i + 5/5j, is this correct?
I feel too lazy to verify that at moment. Just double-check the slope from each graph.

Any time when the particle is moving parallel to the y-axis.
This is where the x vs y graph comes into play. When would the particle by moving parallel to the y-axis in this graph? Hint: tangents.
 
e(ho0n3 said:
I feel too lazy to verify that at moment. Just double-check the slope from each graph.


This is where the x vs y graph comes into play. When would the particle by moving parallel to the y-axis in this graph? Hint: tangents.

the graph looks really weird... it's like a circle
 
  • #10
Good. If it looks like a circle, there is a point where the tangent at that point is parallel to the y-axis. Do you agree?
 
  • #11
yes and that is at point x = 7 and y = 5.. which is at 7.5 sec

how about:

Any time when the particle has come to a stop?
 
  • #12
Hint: The particle is not moving when its velocity is 0.
 
  • #13
I got it now.. thanks for the help :)
 
Last edited:

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