# Velocity from position in a parametric

#### -EquinoX-

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#### e(ho0n3

How about graphing the table of values to start?

#### -EquinoX-

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

#### e(ho0n3

You should have drawn two graphs (why?) and the velocity is the slope (why?).

#### -EquinoX-

the velocity is the slope (why?).

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

why two graph, I don't know

#### e(ho0n3

why two graph, I don't know
You need 2 because you have 2 position components: x vs t and y vs t.

#### -EquinoX-

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.

#### e(ho0n3

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.

#### -EquinoX-

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

#### e(ho0n3

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?

#### -EquinoX-

yes and that is at point x = 7 and y = 5.. which is at 7.5 sec

Any time when the particle has come to a stop?

#### e(ho0n3

Hint: The particle is not moving when its velocity is 0.

#### -EquinoX-

I got it now.. thanks for the help :)

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