For the second part, it looks like a differential equation that I either don't know how to solve or have forgotten how. ma = F + kv^2 Where a, the acceleration, is the second derivative of the position function with respect to time, and v, the velocity, is the first derivative.
I suspect you mean ma = F - kv^2. Air drag makes things slow down, not speed up. Look at the mathworld web site that Zephyr pointed out. It has the answer to your question.
If that were the case he wouldn't have got the tanh solution. That solution arises from m*dv/dt = F - kv^2.
Nonsense results arise if both F and k are positive. In particular, the speed will reach infinity in a finite time. Equations of motion: dx/dt = v m dv/dt = F + k*v^2 with initial states at t=0 given by x=x0, v=v0 Suppose F,k are both positive. Define v_c = sqrt(F/k) t_c = m/sqrt(F*k) u = v/v_c u0 = v0/v_c tau = t/t_cDividing both sides of the velocity equation of motion by v_c yields du/dt = sqrt(k/F)*F/m + sqrt(k/F)*k/m*F/k*u^2or du/dt = 1/t_c*(1+u^2) du/dtau = du/dt*dt/tau = 1+u^2which has solution u(tau) = tan(tau+atan(u0))or v(t) = v_c*tan(t/t_c+atan(v0/v_c)) The velocity goes to infinity as t -> t_c*(pi/2-atan(v0/v_c)) If k is negative, the trigonometric functions become hyperbolic functions by Osborne's Rule -- or just rewrite as m dv/dt = F - k*v^2
Hi HonorAndStrength, Replacing the whole original post is kind of bad form. It ruins the historical record of the discussion. Adding a note after the original question would be better. Something like this: Short question: find the integral of tanh(x + C) with respect to x. Long question: given that the force on an object is F + kv^2, where F is some arbitrary (but constant) force, k is a constant coefficient, and v is the velocity of the object, find the distance the object has travelled after t seconds. Edit: friend needed answer, its been solved now, ty