Deriving Vector and position vectors from Force vector

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Homework Help Overview

The problem involves deriving the position and velocity vectors of an object subjected to a time-dependent force vector. The force components are given in terms of constants and the variable y, raising questions about the integration process and the treatment of variables.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the acceleration, velocity, and position vectors from the given force vector using integration. Some participants question the handling of constants during integration and the implications of treating y as a variable rather than a constant.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach. There is recognition of the need to address the integration of variables correctly, and some guidance has been offered regarding the treatment of y in the force equation.

Contextual Notes

Participants are navigating the complexities of integrating a force that depends on a variable, which may affect the resulting equations for velocity and position. The original poster's assumptions about the constants and variables are under scrutiny.

HclGuy
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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.

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]

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?
 
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You reasoning seems good, but you are forgetting some constants when you integrate.
 
Thanks, just noticed that myself as well.
 
note that in Fx(t) = k1 + k2y, the y is not a constant, so the integral of k2y dt is not equal to k2yt
 

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