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Suppose a particle of mass $m$ was moving under the influence of a velocity dependent retarding force $F(v) = - \alpha \sqrt{v}$, and I wanted to find the acceleration as a function of time, $a(t)$.

Could I just do the following?

$$F(v) = - \alpha \sqrt{v} \implies ma(t) = -\alpha \sqrt{v} \implies a(t) = -\frac{\alpha}{m}\sqrt{v} \qquad (1)$$

This is what is done in one of my lecturers notes. However I think that it must be wrong because if $F(v)$ is a function of time, then surely the acceleration must be also a function of velocity $a(v)$, and thus in turn a function of time as follows $a(v(t))$.

If $(1)$ is indeed correct, could you please explain why it is correct, and why one can do that?

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  • $\begingroup$ While your meaning is reasonably clear, it is formally the case that a constant velocity is still a function of time, simply a very simple one. $\endgroup$ – dmckee May 14 '17 at 2:51
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You just need to solve Newton's second law and it will give the velocity of the particle as function of time. Since acceleration is non vanishing it is clear that velocity is not constant. On the other hand, since acceleration depends on velocity, acceleration is not constant either.

Newton's second law can be written as $$m\frac{dv}{dt}=-\alpha\sqrt v,$$ which shall be integrated as $$\int_{v(0)}^v(t)\frac{dv}{\sqrt v}=-\frac{\alpha}{m}\int_0^t dt.$$ This gives $$\sqrt v=\sqrt{v(0)}-\frac{\alpha}{2m}t.$$ As you can see, velocity is a function of time. Finally, you can plug this result in your Eq. (1), which is correct, to obtain acceleration as a function of time, $$a(t) = -\frac{\alpha}{m}\sqrt{v(0)}+\frac{\alpha^2}{2m^2}t.$$

Such acceleration, depending on the square root of velocity, could be due to a drag force. Suppose the particle was left with some initial velocity and then is under action of this drag force only. The drag will retard the particle and therefore the drag force itself will be time dependent.

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  • $\begingroup$ I saw your answer come in as I was typing mine saving me the work of typing a solution! $\endgroup$ – Alfred Centauri May 13 '17 at 14:31
  • $\begingroup$ @AlfredCentauri I hope you was just in the beginning of typing =) $\endgroup$ – Diracology May 13 '17 at 14:32
  • $\begingroup$ I was! In fact, I had just done a search to find a related previous answer of mine and that's when I saw your answer. $\endgroup$ – Alfred Centauri May 13 '17 at 14:34
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    $\begingroup$ By the way, since $\sqrt{v} \ge 0$, the 3rd equation is valid for $t \le \frac{2m\sqrt{v(0)}}{\alpha}$ and then $v = 0$ otherwise. $\endgroup$ – Alfred Centauri May 13 '17 at 16:03
  • $\begingroup$ @AlfredCentauri That's a good remark. When $t=2m\sqrt{v(0)}/\alpha$, velocity vanishes and then acceleration/force also vanishes. The particle stops after a finite time interval. $\endgroup$ – Diracology May 13 '17 at 16:20
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You can possibly write

F(v)=−α√v as ma(v)=−α√v

a(v)=−α√v/m -----(1)

dv/dt=−α√v/m

dv/√v=−αdt/m integrating,

2√v =−αt/m + c ( apply appropirate limits according to ur question during integration to avoid the integral constant.

so here, √v = (−αt/m + c)/2 -----(2)

according to your question, from (1)

a(v)= −α√v/m

from (2)

a(t)= −α(−αt/m + c)/2m

and suppose the velocity is zero at t=0, then , c=0

Note that: I just took v=0 at t=0 to simplify the equation. But there is no logic in doing so (since you mentioned that F(v) is the only force acting on the body) as F itself is a function of velocity.So if v=0 then F also =0 at t=0. This remains so unless some other force sets the body into motion :)

you get,

a(t)= −α(−αt/m)/2m

which is,

a(t)= α(squared)t/2m(squared)

so here you have the acceleration as a function of time :)

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As you state

$$ a = a(v(t)) = a(t) $$

Perhaps you are thinking something in the lines of $a = a(v, t)$, but then

$$ a = a(v, t) = a(v(t),t) = a(t) $$

That idea here is that $a$ depends explicitly on $v$, but bacause $v$ depends on time, then $a$ depends implicitly depends on $t$

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From the equation \begin{equation} F(v)=-\alpha \sqrt{v}, \end{equation} we can get \begin{equation} ma=-\alpha \sqrt{v} \end{equation} and thus \begin{equation} a(v)=-\frac{\alpha}{m}\sqrt{v}, \end{equation} which means that the acceleration is a function of velocity. But $a(v)\neq a(t)$ because we don't know whether $v$ is time-dependent or not ($v$ can be constant and thus time-independent). So we can't get $a(t)=-\frac{\alpha}{m}\sqrt{v}$ in general.

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    $\begingroup$ "$v$ can be constant and thus time-independent" - careful... this is true only if $v = 0$. If $v\ne 0$, then the acceleration is non-zero and thus, the velocity is not constant. $\endgroup$ – Alfred Centauri May 13 '17 at 14:11
  • $\begingroup$ No. If $v\neq 0$, acceleration can also be zero as long as $v= constant$. $\endgroup$ – Shen May 13 '17 at 14:54
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    $\begingroup$ No, if $v$ is constant, than $a$ is zero (by definition) but if $a$ is zero, it must be that $\sqrt{v} = 0$ (by the equation given). It's as plain as that. $\endgroup$ – Alfred Centauri May 13 '17 at 15:05
  • $\begingroup$ If v $ \not = 0 $, then a is non-zero and since a is the rate at which v changes, v $ \not = 0 $ implies v must change under the influence of the mentioned force so that if v $ \not = 0 $, then the velocity is not constant. $\endgroup$ – PhyEnthusiast May 16 '17 at 10:30
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You've stumbled onto one of the most common confusions in classical mechanics, though it doesn't usually come up until a second course: you can view everything as either a function of a configuration, or a function of a trajectory.

When they say $F = F(v)$, they mean that a force could generally depend on everything about a particle's current configuration, which in this case is its position and velocity, but in this particular case, the force only happens to depend on the current velocity.

However, when we're talking about the trajectory of a particle, everything can be written as a function of time $t$; you can always ask what the force, velocity, etc. are at a specific time. Moreover you can define anything to be locally a function of anything else. For example, I could invert $x(t)$ to get $t(x)$, and then write $v$ as a function of $x$ through $v = v(t(x))$.

In the former case the function arguments are essential; in the latter they can be almost whatever you want. Typically we use the former when setting up equations and the latter when solving them, and your lecturer just switched without mentioning it.

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