Time until it stops moving

Nov 2019
15
1
Athens
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?
 
Jun 2016
1,151
520
England
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?
 
Jun 2016
1,151
520
England
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.
 
Nov 2019
15
1
Athens
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.
 
Jan 2019
33
26
$v = v_0 - \dfrac{bx^2}{2m}$

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.
 
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Nov 2019
15
1
Athens
$v = v_0 - \dfrac{bx^2}{2m}$

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 !!!!!!!!
 
Jun 2016
1,151
520
England
Oops (again)
I should have twigged that there was a rate of change of acceleration
(what a Jerk).
 
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