• Support PF! Buy your school textbooks, materials and every day products Here!

Deriving Vector and position vectors from Force vector

  • Thread starter HclGuy
  • Start date
13
0
1. Homework Statement
An object of mass m is at rest at equilibirum at the origin. At t=0, a new force [tex]\vec{F}(t)[/tex] is applied that has components
[tex]F_{x}[/tex](t) = k1+k2y [tex]F_{y}[/tex](t)=k3t
where k1, k2, and k3 are constants. Calculate the position r(t) and velocity v(t) vectors as functions of time.
2. Homework Equations
We know that Force = ma.
and that :
[tex]\int \vec{a}dt = \vec{v}(t)[/tex]
[tex]\int \vec{v}dt = \vec{r}(t)[/tex]

3. The Attempt at a Solution

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

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

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

Answers and Replies

G01
Homework Helper
Gold Member
2,649
16
You reasoning seems good, but you are forgetting some constants when you integrate.
 
13
0
Thanks, just noticed that myself as well.
 
25
0
note that in Fx(t) = k1 + k2y, the y is not a constant, so the integral of k2y dt is not equal to k2yt
 

Related Threads for: Deriving Vector and position vectors from Force vector

Replies
0
Views
6K
Replies
9
Views
14K
  • Last Post
Replies
3
Views
679
Replies
3
Views
34K
Replies
1
Views
478
  • Last Post
Replies
8
Views
43K
  • Last Post
Replies
4
Views
4K
Replies
0
Views
1K
Top