How can air resistance be incorporated into the falling particle equation?

AI Thread Summary
The discussion focuses on incorporating air resistance into the falling particle equation, specifically examining two cases: linear air resistance (av) and quadratic air resistance (Bv^2). The first part was solved by integrating the equation of motion, leading to a relationship between velocity and distance fallen. For the quadratic case, participants suggest using the terminal velocity substitution to simplify the equation, allowing for separation of variables and integration. Alternative methods, such as partial fraction decomposition, are also recommended for solving the integrals involved. Overall, the thread emphasizes the importance of different mathematical techniques in tackling the problem of air resistance in falling particles.
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"A particle is relased from rest (y = 0)and falls under the influence of gravity and air resistance. Find the relationship between v and the distance of falling y when the air resistance is equal to a) av and b) Bv^2."

I already solved part a by integrating dv/dt = -g - (a/m)(dy/dt) with respect to t to give me v = -gt - (ay/m), and then taking force = m(dv/dt) = -mg-av and rearranging this to give me dv/(g + av/m) = -dt, which I integrated to get (m/a)ln(av/m + g) = -t + c. Solving for c and doing some algebra gives me y = -(m/a)(v -(mg/a)ln(av/mg + 1)) as my final solution for y.

Part b, though, I have no idea how to do. If I try and solve it the same way, I end up with the integral of dv/(g + av^2/m) with respect to time, which I have no idea how to solve. I also have the integral -g-(av^2)/m with respect to time, which I don't know how to solve either. If anyone can help me solve these integrals, or point out an easier way to solve this problem, I'd be most grateful.
 
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Does anyone know how to do this? I really need help on this one and it's due tomorrow.
 
ok, you have

<br /> m\dot{v}=mg-bv^2<br />

this will be a lot easier to do if you make the substitution

<br /> v_t =\sqrt{\frac{mg}{b}}<br />

where v_t is the terminal velocity which is when the right hand side of the eq. balances out. subing in v_t we get

<br /> <br /> \dot{v}= g(1-\frac{v^2}{v_t^2})<br />

then from here you can use separation of varbs to get

<br /> <br /> \frac{dv}{1-\frac{v^2}{v_t^2}}=gdt<br />

and simply integrat over v and t.
it may be very helpful to look at the hyperbolic functions most notably that of the arctanh(x).
 
Thanks man. That helps a lot. Now if only I'd studied hyperbolic functions...
 
You don't need hyperbolic functions, although that is an alegant way to do it. Another option is to factor the denominator into a sum and difference and then apply partial fraction decomposition. The resulting fractions only have linear denominators and can easily be integrated.
 
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