What is the velocity, speed, position and distance traveled after 30 minutes?

[xxChrisxx]

I understand that (z=up, y=toward you, x=across) I just don't understand how to apply it to the equations and which equations to use.

Well let's apply that to the situation. Forget the equations for the moment, and just think about what's going on.

You said it's a balloon,
so it can move north/south. We'll call that X.
With south in the +ve direction.

East/west we'll call that Y.
East in the +ve direction.

Up/down we'll call that Z.
Up is +ve.So the balloon can move in 3 different axis. In reality it can move in any direction. But that real movement can be defined as though you walked up/down, east/west, north/south.Do you know what resultants and components of something are?edit: I'm leaving work now. if you don't know what they are read the following:
http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html
http://www.mathwarehouse.com/vectors/resultant-vector.php

 

[physics(L)10]

Isn't Up and down the same thing as north and south?

resultant is formed when you add/subtract two vectors together. components I probably know but I don't know they're called components so I don't know lol.

edit: component is a particular vector ex. (-1,10)?

 

[physics(L)10]

I've been working on it and I think I mightve gotten something:

r=(0,0,0)
u=(0,0,0)
a=(10,5,10)

and then you use the equation x=Xo +ut + at^2/2 for x,y,z and you get 3 velocity values. Now I'm stumped

 

[xxChrisxx]

Isn't Up and down the same thing as north and south?

Up and down is elevation.

resultant is formed when you add/subtract two vectors together. components I probably know but I don't know they're called components so I don't know lol.

and then you use the equation x=Xo +ut + at^2/2 for x,y,z and you get 3 velocity values. Now I'm stumped

Yep you should have 3 velocity values, and three displacement values.

Look at this graph. There are three pink lines, one going down x, along y and then up z. Those three that you've calculated correspond to those lines. Also you can see that the three displacements give you the new position (x,y,z).

What you now want to do is find the resultant of those three vectors. So that the path taken takes you straight from (0,0,0) to (x,y,z). #This is what you found originally.

 

[kuruman]

I've been working on it and I think I mightve gotten something:

r=(0,0,0)
u=(0,0,0)
a=(10,5,10)

and then you use the equation x=Xo +ut + at^2/2 for x,y,z and you get 3 velocity values. Now I'm stumped

What do the equations that you have written down represent? If you want to write the three vectors at time t = 0, you should write

r=(0,0,0)
u=(10,0,0)
a=(10,5,10)

Note that the x component of the initial velocity is 10 m/s as given by the problem. Now can you write the position, velocity and acceleration vectors not at just t = 0 but at any time t? Hint: The acceleration vector will be the same. What about the other two?

 

[physics(L)10]

Alright, thanks guys I think I got it.