- #1

Matty R

- 83

- 0

I was hoping someone could help me with this mechanics problem.

## Homework Statement

A particle of 2kg mass moves in three dimensions under the action of a force [tex]\underline{F}=6\underline{i}-6t^2 \underline{j}+(4-2t)\underline{k}[/tex].

At [tex]t=0[/tex], it is at rest with position vector [tex]\underline{r}=\underline{j}[/tex]

Find the velocity and position vectors as functions of time.

## Homework Equations

[tex]\underline{F}=m \underline{a}[/tex]

[tex]\underline{a}=\frac{d \underline{v}}{dt}[/tex]

[tex]\underline{v}=\int(\underline{a})dt[/tex]

[tex]\underline{v}=\frac{d \underline{r}}{dt}[/tex]

[tex]\underline{r}=\int(\underline{v})dt[/tex]

## The Attempt at a Solution

VELOCITY:

[tex]\underline{F}=m \underline{a}[/tex]

[tex]\underline{a}=\frac{\underline{F}}{m}[/tex]

[tex]=\frac{6\underline{i}-6t^2 \underline{j}+(4-2t)\underline{k}}{2}[/tex]

[tex]=3\underline{i}-3t^2 \underline{j}+(2-t)\underline{k}[/tex]

[tex]\underline{a}=\frac{d \underline{v}}{dt}[/tex]

[tex]\underline{v}=\int(\underline{a})dt[/tex]

[tex]=\int(3-3t^2+(2-t))dt[/tex]

[tex]=(3t+c_1)\underline{i}-(t^3+c_2)\underline{j}+(2t-\frac{1}{2}t^2+c_3)\underline{k}[/tex]

I don't know what to do with the constants. I keep getting them all as 0, unless that is the right thing to do.

POSITION:

[tex]\underline{v}=\frac{d \underline{r}}{dt}[/tex]

[tex]\underline{r}=\int(\underline{v})dt[/tex]

[tex]=\int(3t-t^3+(2t-\frac{1}{2}t^2))[/tex]

[tex]=\frac{3}{2}t^2 \underline{i}-\frac{1}{4}t^4 \underline{j}+(t^2-\frac{1}{6}t^3) \underline{k}+ \underline{j}[/tex]

[tex]=\frac{3}{2}t^2 \underline{i}+\frac{3}{4}t^4 \underline{j}+(t^2-\frac{1}{6}t^3) \underline{k}[/tex]

I find mechanics very interesting, but I also find it quite confusing.

Could anyone help me please?