Why does resisting force depend on velocity? I think there is no relation between resisting force and velocity of object. Please speak about it logically.
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The air resistance, a force sometimes known as a drag, always increases with the speed but the precise dependence on the speed differs in various situations, especially depending on the presence of turbulence etc.
For low speeds, the air resistance is proportional to the velocity, $\vec F = -b\vec v$. This is known as Stokes' drag. A direct proportionality is the only natural, smooth function of a vector mapped to another vector. The unit (or constant-length) vector in the direction of motion isn't natural as it would be discontinuous for $\vec v =0$.
When you're moving by speed $v$, a higher number of particles hit you on the front side. The difference between collisions from the front and from the back go like $v$ and because each collision with a (very fast) particle effectively changes your momentum by the same amount, the change of the momentum per unit time – the force – will be proportional to the velocity, too.
At higher velocities, the drag increases as a higher power, typically quadratic power, of the velocity. That's for a higher Reynolds number, for turbulence. It takes some effort to quickly explain the scaling but it can be done. Well, let me say something: $$ F_D\, =\, \tfrac12\, \rho\, v^2\, C_d\, A, $$ This is the formula for the drag. It doesn't depend on the velocity at all (because the drag is due to universal turbulence and "vortices" and not just some surface friction). Because both sides have to have the same units, $F$ has units of mass times distance per squared time (mass times acceleration), the velocity is the only factor that can give us powers of a second (the unit of time), and it has to be squared to produce the right power. We call this argument "dimensional analysis".
Luboš Motl
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