linear air drag

QUESTION:
If you have two objects with exactly the same volume and surface area, and if you apply an equal force on both of them, the one with less mass will have a greater acceleration. This is because F=ma. But, if you do this experiment in any fluid, which one will go farther? The surface areas are the same so neither will have a greater form drag. But because acceleration due to force, the object with less mass will slow down much more than the other It is also known that the drag acting on an object is directly proportional to the square of its speed. So, even if the object with less mass has a greater acceleration in the beginning, it will slow down much more than the one with greater mass. So which one will travel farther?

NOTE ADDED:
After several communications, the question is this: If the same force F acts on both objects for the same time and then stops, which object goes farther? There is no gravity, so this is a one-dimensional problem. Both objects start at rest. This is one of those quite technical answers although, without understanding all the math, you can see the qualitative behavior. This is going to be a lengthy answer, so maybe you just might want to skip it!

ANSWER:
I will assume that the drag force is given by F_drag=-bv where b is the same for both objects (as stipulated by the questioner).

For the first part of the problem, Newton's second law is mdv/dt=F-bv. At large times the velocity will become essentially equal to a constant because F≈bv; this speed is called the terminal velocity v_t=(F/b). Without going into any detail the solution is v=v_t (1-e^-t/^τ) where τ=m/b is the characteristic time. This is plotted on the first figure above; the larger the mass, the longer it takes to reach the terminal velocity so, as noted by the questioner, the lighter mass will go farther in a given time. To get the position as a function of time, we need to solve v=v_t (1-e^-t/^τ); the solution is x=v_t[t-τ(1-e^-t/^τ)]. This is plotted in the second figure above; the smaller mass goes farther but after a long enough time, the two keep a constant distance apart.
For the second part of the problem, Newton's second law is mdv/dt=-bv. Now, the velocity will begin with a speed v₀ which is the speed at the end of the time interval in the first part. Without going into any detail the solution is v=v₀e^-t/^τ. This is plotted on the third figure above; the smaller mass clearly loses velocity more quickly as expected. To get the position as a function of time, we need to solve v=dx/dt=v₀e^-t/^τ; the solution is x=v₀τ(1-e^-t/^τ). This is plotted in the fourth figure above; if the two start with equal speeds at the same place (which they do not), the larger mass would always be ahead of the smaller mass. Here, unlike the quadratic drag case, the objects reach a maximum distance and so we do not need to make approximations regarding how far each goes. It now turns out that the problem can be done analytically quite easily. The maximum distance for the second part of the problem is for t=∞, so x₂=v₀τ with v₀=v_t(1-e^-t/^τ) where t is the time in the first part until the force is stopped. Now, x₁=v_t[t-τ(1-e^-t/^τ)] and so x₁+x₂=v_t[t-τ(1-e^-t/^τ)]+v_tτ(1-e^-t/^τ)=v_tt. The distance traveled is independent of the mass.