How Does Air Resistance Affect the Velocity of a Falling Object?

AI Thread Summary
Air resistance affects the velocity of a falling object by introducing a force that opposes gravity, expressed as R = kv². The discussion focuses on deriving the speed of a falling ball in terms of the distance fallen, starting from the equation ma = mg - kv². The integration process is outlined, with emphasis on separating variables correctly to express acceleration in terms of velocity and distance. The terminal velocity is identified as v = √(g/k(1-e²kx)), highlighting the importance of checking units and limits in derived formulas. Overall, understanding the relationship between air resistance and falling velocity is crucial for accurate modeling of motion.
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A ball of mass m is released from rest at x=0. Air resistance is expressed as R = kv2 , where k is a positive constant and v denotes velocity.

Derive an expression for the speed in terms of the distance x that it has fallen.
Identify the terminal v.

I know I have to start with ma = mg - kv2 and integrate but I can't seem to separate the variables right to get it as function of x or distance (even though the problem has a picture of a vertical fall)
 
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Hint: Express the 'a' in your equation differently.
 
Okay so I have mv(dv/dx) = mg - kv2 , and divide both sides by the (mg - kv2)

Then, (mv/(mg - kv2))dv = dx

∫(mv/mg - mv/kv2) dv = ∫dx

Just need a little help integrating if I'm on the right track I guess...
 
Acceleration is not the derivative of velocity with respect to distance; it is the derivative with respect to time.
 
Great...just realized that
 
If I change a to d2s/dt2 is this the right path?
 
Since your force is given in terms of v, it would be better to write the acceleration as dv/dt:
m\frac{dv}{dt}= mg- kv^2
and there is no problem with "separating variables"- you have
\frac{mdv}{mg- kv^2}= dt
 
But I need the velocity in terms of the distance fallen, x. Integrating the right side, ∫dt , would just give me 't' am I right?
 
Was I not correct in saying that a = v (dv/ds) ??
 
  • #10
In my book it says v dv = a ds --> a = v(dv/ds)...

so I have vdv/(-g + kv2) = dx

∫vdv/(-g + kv2) [from 0 to v] = ∫dx [from 0 to x]

(1/2k) ln(-g+kv2) [from 0 to v] = x

--> (1/2k) ln(g-kv2/g) = x

Solving for v and skipping a few steps I got v = √(g/k(1-e2kx))
 
  • #11
Your units do not show velocity. You need a mass in there.
 
  • #12
You're right...should start as dx = mvdv/(-mg + kv2)
 
  • #13
Whenever you derive something, it is always prudent to check your units in the final product to see that they are what is intended. Secondly, it is a good idea to explore the limits of your formula to see if it makes sense at its limits.
 
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