# Time until it stops moving

#### Maidenas

Object with mass "m" is moving in an horizontal plane surface with initial velocity of magnitude $$\displaystyle v_0$$.
The object is subject to a friction force with magnitude $$\displaystyle F= bxv$$ where b is positive constant, x is the space the object has travelled and $$\displaystyle v$$ its speed.

a) Find the space the object travelled until it stops moving
b) Find the time it needs the ohject to stop moving.

The 1st question i did it. After
I found that $$\displaystyle x = \sqrt{ \frac{2mυ_0}{b}}$$ m the object will stop moving

Also i found that $$\displaystyle v=- \frac{bx^2}{2m}+ v_0$$

How i will find the time it need the object from stop moving?

#### Woody

Your starting point has to be:
$$\displaystyle s=ut+1/2at^2$$
and:
$$\displaystyle v=u+at$$

you have:
$$\displaystyle F=bxv$$ and $$\displaystyle F=ma$$
thus:
$$\displaystyle a=bxv/m$$
so with $$\displaystyle u$$ being the initial velocity and $$\displaystyle s$$ being the displacement (so $$\displaystyle s=x$$).

can you go on from here?

#### Woody

These equations giving the relationship between Force, Mass, Acceleration, Velocity and Displacement are well worth remembering.
They come up time and again on this Forum, and are a favorite of exam writers.
They allow a question to be put together that looks complicated and frightening, because it has a lot of moving parts
(sorry the pun was just too obvious to resist)
But if you remember these equations, you will find that, what looks like a difficult question, should break down into several steps, each of which is actually quite easy.

#### Maidenas

Ι think that thε first two equations are for uniformly accelerated motion. This excersice doesn't say anything about the kind of motion the object is doing.

#### Cervesa

$v = v_0 - \dfrac{bx^2}{2m}$

let $p^2 = v_0$ and $q^2 = \dfrac{b}{2m}$, both constants

$\dfrac{dx}{dt} = p^2 - q^2x^2$

$\dfrac{dx}{(p-qx)(p+qx)} = dt$

the method of partial fractions yields

$\dfrac{1}{2p}\left(\dfrac{1}{p+qx} + \dfrac{1}{p-qx} \right) \, dx = dt$

$\left(\dfrac{1}{p+qx} + \dfrac{1}{p-qx} \right) \, dx = 2p \, dt$

$\left(\dfrac{q}{p+qx} - \dfrac{-q}{p-qx} \right) \, dx = 2pq \, dt$

$\log{\left|\dfrac{p+qx}{p-qx}\right|} = 2pq \cdot t + C$

I leave the task of solving for $x$ as a function of time and back-substituting for $p$ and $q$ to you.

topsquark

#### Maidenas

$v = v_0 - \dfrac{bx^2}{2m}$

let $p^2 = v_0$ and $q^2 = \dfrac{b}{2m}$, both constants

$\dfrac{dx}{dt} = p^2 - q^2x^2$

$\dfrac{dx}{(p-qx)(p+qx)} = dt$

the method of partial fractions yields

$\dfrac{1}{2p}\left(\dfrac{1}{p+qx} + \dfrac{1}{p-qx} \right) \, dx = dt$

$\left(\dfrac{1}{p+qx} + \dfrac{1}{p-qx} \right) \, dx = 2p \, dt$

$\left(\dfrac{q}{p+qx} - \dfrac{-q}{p-qx} \right) \, dx = 2pq \, dt$

$\log{\left|\dfrac{p+qx}{p-qx}\right|} = 2pq \cdot t + C$

I leave the task of solving for $x$ as a function of time and back-substituting for $p$ and $q$ to you.
My friend thank you very much for the help !!!!!!!!

#### Woody

Oops (again)
I should have twigged that there was a rate of change of acceleration
(what a Jerk).

topsquark

Forum Staff