# Finding the diferencial function of a moving particle

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1. Nov 9, 2015

### Gbox

1. The problem statement, all variables and given/known data
A particle is moving in a liquid $a=-kv$ when $k$ is a constant, if a constant force is being apply on the particle $a=-kv+\frac{F}{m}$.
1. for the formula thats include the force, find the diferencial function where $v$ will appear in it explicitly, ad its derivites
2. define a new variable $U=v-\frac{F}{mk}$ and substitute in place of $v$ that is in the diferencial function.
3. guess a solution in the form of $U=AE^{-Bt}$ where A and B are contestants, find B that for it the equation is true
4. find A and when the velocity at $t=0$ is 0
5.find $x(t)$ when $x(t=0)=0$
6. according to the result on 4, what will be the velocity of the particle when $t\rightarrow \infty$
2. Relevant equations
$a=-kv$
$x+x_o+v_0t+\frac{at^2}{2}$

3. The attempt at a solution
1. I should be looking for the functions which its derivites is $a=-kv+\frac{F}{m}$ so $v=-kvt+ frac{Ft}{m}$
And $x=x_0+\frac{kvt^2}{2}+\frac{Ft^2}{2m}$?

2.$x=x_0+\frac{k*(v-\frac{F}{mk})*t^2}{2}+\frac{Ft^2}{2m}$

Are 1 and 2 are correct?

2. Nov 9, 2015

what is a

3. Nov 9, 2015

### Gbox

a Is the acceleration

4. Nov 9, 2015

### Let'sthink

Statement of the question is in a clumsy language and confusing. Moment we say a particle is moving with a = -kv, means force is acting on it = ma = -mkv. and this is a velocity dependent force. Now if additional constant force is applied, we have new acceleration a' given by
ma' = F - kmv or
a' = d²x/dt² = (F/m) - k(dx/dt), because we want an expression with v and dv/dt, we have
dv/dt = (F/m) - kv
Your very first step is wrong. Integral of kv is not kvt because v is not a constant but function of t in general in both cases when F = or F is non-zero.. The rest thing you try using your mathematical skills related to integral calculus.