Velocity from position in a parametric

[-EquinoX-]

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

 

[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

how about:

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 :)