friend needed answer, its been solved now, ty

What have you done so far?

Will you need to know the mass?

Which of Newton's 'laws' do you think might be useful in part 2?

What's tanh(x+C)? Is that supposed to be tan(x+C), or the tangent of [h(x+C)]?

tanh is the "hyperbolic tan" function.

http://mathworld.wolfram.com/HyperbolicTangent.html

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.

tanh is the "hyperbolic tan" function.

http://mathworld.wolfram.com/HyperbolicTangent.html


Ugh.
Nevermind then.

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.

It could be correct, if it's an object being slowed by reverse thrust + parachute, or something.

It could be correct, if it's an object being slowed by reverse thrust + parachute, or something.

If that were the case he wouldn't have got the tanh solution. That solution arises from m*dv/dt = F - kv^2.

Hey, I didn't come up with the formula! heh.

Hey, I didn't come up with the formula! heh.

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_c
Dividing 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^2
or
du/dt = 1/t_c*(1+u^2)
du/dtau = du/dt*dt/tau = 1+u^2
which 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

edited : fixed now

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