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Velocity from position in a parametric

  • Thread starter -EquinoX-
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How about graphing the table of values to start?
 
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
 
1,356
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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
 
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.
 
1,356
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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.
 
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
 
1,356
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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?
 
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?
 
1,356
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Hint: The particle is not moving when its velocity is 0.
 
I got it now.. thanks for the help :)
 
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