7
$\begingroup$

Is it possible that acceleration may vary linearly with velocity. Is it practically possible, if so is there a practical example of it?

By integration I was able to verify that for the above case to hold true, we need velocity to depend linearly on position, but the question remains there that is it practically possible or it is a mere theoretical situation?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ A trivial example is a stationary object. $\endgroup$
    – mmesser314
    Commented Apr 20, 2021 at 13:42
  • 3
    $\begingroup$ If you're asking for an acceleration in the direction of velocity, which increases as velocity increases, there are practical reasons why things in the real world don't work that way, at least not for very long :) $\endgroup$
    – hobbs
    Commented Apr 20, 2021 at 23:40
  • $\begingroup$ Are you asking about how to solve the differential equation $\ddot{x} = k \dot{x}$? $\endgroup$
    – Prof. Legolasov
    Commented Apr 23, 2021 at 11:43

6 Answers 6

Reset to default
18
$\begingroup$

Yes, air resistance at low velocities is one such example, but I'm sure that there are others. For an object moving at a suitably low velocity, the drag on the object is given by $$\vec{F} = - b \vec{v} ,$$ which of course means that the acceleration is varying linearly with the velocity. This is the sort of drag experienced by an object moving slowly in a laminar fluid, i.e. when we have very low Reynolds number.

Improve this answer
$\endgroup$
10
$\begingroup$

For a charged ($q$) particle ($m$) in a magnetic field ($B_i$), the acceleration ($a_i$) is:

$$ a_i = \frac 1 m F_i= \frac 1 m [-q\epsilon_{ijk}B_jv_k ]$$

or explicitly as a linear operator:

$$ a_i =\big[\frac{-q} m\epsilon_{ijk}B_j\big]v_k$$

Improve this answer
$\endgroup$
5
  • $\begingroup$ This holds only for non-relativistic particles, but still nice example. $\endgroup$
    – user1079505
    Commented Apr 21, 2021 at 1:23
  • 1
    $\begingroup$ This is of course perpendicular to the direction of motion, so it might not be matching OP's idea of "we need velocity to depend linearly on position". Although I guess you could say that the direction of the velocity vector does vary just with position, but I'm not sure I'd describe an orbit as "linearly" with position. $\endgroup$
    – Peter Cordes
    Commented Apr 21, 2021 at 8:43
  • $\begingroup$ @PeterCordes in this example acceleration depends linearly on velocity because $-\frac{q}{m} \epsilon_{ijk} B_j$ is a linear operator. It is rotation, but rotations are linear. $\endgroup$
    – Charles Hudgins
    Commented Apr 21, 2021 at 8:47
  • $\begingroup$ @CharlesHudgins: And integrating, as mentioned in the question, will give velocity linearly dependent on position? My calculus is slightly too rusty to be sure, which is why I asked. And even if it's linear in that sense, I still wonder if the querent was thinking of straight-line motion, and wanted to point out that basic fact. $\endgroup$
    – Peter Cordes
    Commented Apr 21, 2021 at 8:56
  • 1
    $\begingroup$ @PeterCordes If you integrate, you will find that velocity has an affine linear dependence on position, i.e. $\vec{v} = A \vec{x} + \vec{b}$ for some linear operator $A$ and some constant vector $\vec{b}$. If the OP thought that linear meant straight line motion, then this is a good example for clarifying some vocabulary. "Acceleration varies linearly with velocity" means there exists a linear operator $A$ such that $\vec{a} = A \vec{v}$. Linear motion means $\vec{x}(t) = f(t) \vec{b} + \vec{c}$ for some function $f(t)$ and vectors $\vec{b}$ and $\vec{c}$. $\endgroup$
    – Charles Hudgins
    Commented Apr 21, 2021 at 9:48
6
$\begingroup$

One of the basic elements (along with masses and springs) used to model mechanical systems is the damper, which provides force equal to the relative velocity of its two ends.

Real world implementations are dashpots and hydraulic or pneumatic dampers use in vehicle suspensions, aka shock absorbers or gas struts.

Improve this answer
$\endgroup$
3
$\begingroup$

It is certainly possible for force to vary linearly with velocity. You could easily construct a scenario to make this happen. The easiest way to construct such a scenario would be with an electric motor where you control the power based on the velocity.

Since $P=F\cdot v$ then if $F=kv$ we have $P=kv^2$. So if you simply monitor your velocity and deliver power at a rate $P=kv^2$ then you would have a force linearly related to velocity.

Improve this answer
$\endgroup$
3
$\begingroup$

Consider the ideal dash-pot or the real-life damper. These devices produce a resistance force proportional to speed.

$$ F =- c\, v $$

Coupled with a spring and a mass, it produces the damped spring-mass system that is a very common device.

The force above is the result of oil squeezing through a narrow passageway (usually the outside of a cylinder) with very laminary flow. The governing process is called Couette Flow and produces a linear relationship between shear stress in the fluid and velocity gradient. The gradient is proportional to the speed on one side of the damper, and the shear stresses proportional to the applied force.

Improve this answer
$\endgroup$
1
$\begingroup$

I like your question because, besides the technical side, it gives us a good example of one of the many ways mathematics and physics interact. What you are asking is if the ordinary differential equation $\frac{d^2 x(t)}{dt^2} = k \frac{dx(t)}{dt}$, ($k$ a real constant) with some kind of initial conditions, has a physical meaning in Newtonian mechanics. If we use, for example, $x(0) = 1$ and $v(0) = 1$ we can see that the solution exists and it is an exponential. Now the question comes: does this exponential solution represents a valid Newtonian physical movement of a particle? At least as far as $k > 0$ i think the answer is yes, as the previous answers have shown for concrete cases.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.