Imagine that an object of mass m is initially moving in the x direction through a viscous fluid at a speed v. If the only force acting on this object is a viscous drag force whose magnitude is given by F=bv, prove that the object's x-velocity is given by vx(t)=ve^-qt and determine how the constant q depends on b and m. (this problem requires some calculus involving exponentials and logarithms)

Question

anonymous · Answer

Hint: This is actually a motion-from-force problem. Divide both sides of the x component of Newton's second law by v, and take the indefinite integral of both sides

paxpolaris · Answer

$$F=bv$$$$\implies{ma}=bv$$$$\implies{ma \over  v}=b$$

paxpolaris · Answer

$$\int\limits {ma \over v} dt=bt$$

anonymous · Answer

I think PaxPolaris can help you out with this question better than i can. But thanks for thinking i can help. Im really good at algebra or math so if you need help with that. than give me a call :) good luck

paxpolaris · Answer

dv/dt = a

so  $dv = a\ dt$

paxpolaris · Answer

$$\large \implies\int\limits \frac mvdv=bt$$

anonymous · Answer

So, b/m is equivalent to q

anonymous · Answer

I got v=e^(bt/m)

paxpolaris · Answer

$$m \ln v = bt + C$$

-b/m is equivalent to q

anonymous · Answer

C=Vinitial

paxpolaris · Answer

$$\large v_x(t)=v_{initial} \cdot e^{bt/m} ...$$   

$$v_{initial}=e^{C/m}$$ ... but that's not relevant to the question.

$$-q= b/m \ \implies q=-b/m$$

anonymous · Answer

Awesome, that makes so much more sense now. Thank you for all your help!