# Deriving Vector and position vectors from Force vector

1. Apr 18, 2008

### HclGuy

1. The problem statement, all variables and given/known data
An object of mass m is at rest at equilibirum at the origin. At t=0, a new force $$\vec{F}(t)$$ is applied that has components
$$F_{x}$$(t) = k1+k2y $$F_{y}$$(t)=k3t
where k1, k2, and k3 are constants. Calculate the position r(t) and velocity v(t) vectors as functions of time.
2. Relevant equations
We know that Force = ma.
and that :
$$\int \vec{a}dt = \vec{v}(t)$$
$$\int \vec{v}dt = \vec{r}(t)$$

3. The attempt at a solution

I'm not sure if I'm doing this right but
I did
$$\vec{F}(t)$$ =$$(k1+k2y)\hat{i}$$+$$(k3t)\hat{j}$$
I divided the Force vector by the scalar value of m, the mass to get $$\vec{a}$$
$$\vec{a}(t)$$ = $$(k1+k2y)/m\hat{i}$$+$$(k3t)/m\hat{j}$$

$$\vec{v}(t)$$=$$\int \vec{a}dt$$ = $$(k1+k2y)t/m \hat{i}$$ + $$(k3t^2)/2m \hat{j}$$

then integrate the velocity vector to get the position vector, am I doing this right at all?

2. Apr 18, 2008

### G01

You reasoning seems good, but you are forgetting some constants when you integrate.

3. Apr 18, 2008

### HclGuy

Thanks, just noticed that myself as well.

4. Apr 19, 2008

### SimonZ

note that in Fx(t) = k1 + k2y, the y is not a constant, so the integral of k2y dt is not equal to k2yt