# Clarification of 3D Vectors

1. Sep 25, 2007

### fubag

An object has an initial velocity given by V = vi + vj, where v=10m/s. If an acceleration of A = ak, where a = 10 m/s^2 is applied for 10 seconds determine the final velocity for the object.

How far did the object travel in 10s?

How far from the z axis is the object?

Well to start. I used the equation:
Vf = Vi + at
Vf = (10i + 10j) + (10k)(10s)
Vf = 10i + 10j + 100k

Then to determine distance, I used:
D^2 = (change in x)^2 + (change in y)^2 + (change in z)^2
so.

D = sqrt.(10,200) m?
I am not sure if I calculated this right.

Then for the last part of the question, I am not really sure how to determine this. I know that the object moved 100m in the x direction, and 100m in the y direction, and I think it moved 500m in the z direction given the Vf= 100m/s in z direction and Vi = 0m/s in z direction.

2. Sep 25, 2007

### Leong

for the second part, i think you should use
s = ut + half*at^2 to find its position vector and then use the distance formula to find its distance from the origin.

for the third part, that will be the z-component of the above position vector that you find.

3. Sep 25, 2007

### fubag

so if I use that equation:

i get:

s(t) = 100i + 100j + 500K

D = sqrt. (270,000) m?

Then for distance from z axis is simply sqrt. (500) m?

4. Sep 25, 2007

### Leong

i did a mistake here.

for the second part, i guess so...

for the third part, it is square root of (x-component^2+y-component^2) = sqrt(10000+10000). if you draw the final coordinate of the particle, i think you will see why.

5. Sep 25, 2007

### fubag

i am not really understanding how you determined the 3rd part...

The way you did it, it simply looks like the displacement between the x and y components...not the distance from the z axis.