# Vectors, Check my work.

1. Feb 6, 2009

### Kaln0s

1. The problem statement, all variables and given/known data

A particle of mass 2.0 kg has position given by the vector r = (t^4 -3t^3) i + (t^3 - 9t) j + (2t -6) k
(where i, j, and k are the unit direction vectors and r is given in meters and t in seconds. At what time in seconds is the particle located at the origin?

For this I get t = 3(seconds).

It's the second part I want you to check.

1. what is the magnitude of the particle's velocity (in m/s) when the particle is at the origin?

2. For the particle in question 1, what is the x/y/z-component of the net force (in Newtons) acting on it when it is at the origin? Do not write Newtons as part of your answer.

2. Relevant equations

V = Dr / Dt
F= Ma

3. The attempt at a solution

1. Derive each component individually. Got:

(4t^3-9t^2)i, (3t^2 - 9)j, (2)k
Plug in 3 to get. 27 + 18 + 2 = Sqrt 47 = 6.86 m/s while at origin.

2. Derived each component again to get the acceleration eq.
I got (12t^2 - 18t)i, (6t)j, and (0)k

Then did F = MA for each component.

2.0(12(3)^2 - 18(3)) = 108 for x component.
2.0(6(3)) = 36 for the y component.
2.0(0) = 0 for the z component.

Did I do this correctly?

Thanks for helping :D.

2. Feb 6, 2009

### gabbagabbahey

Hi Kaln0s, welcome to PF!

Don't you have to square each component before summing them up and taking the square root?

$$||\vec{v}||=\sqrt{v_x^2+v_y^2+v_z^2}$$

Looks good to me

3. Feb 6, 2009

### Kaln0s

Oh god I feel dumb as hell lol. I've been doing Calculus and physics non-stop for the past day -_-.

27^2 + 18^2 + 2^2 = 1057 sqrt = 32.5 m/s
:rofl:

Thank you very much.