Can acceleration depend linearly on velocity? 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?
 A: 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.
A: 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.
A: 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.
A: 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.
A: 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:
$$ a_i =\big[\frac{-q} m\epsilon_{ijk}B_j\big]v_k$$
A: 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.
