Posted by **Joe** on Tuesday, September 8, 2009 at 2:42am.

Air resistance acting on a falling body can be taken into account by the approximate relation for the acceleration:

a=dv/dt=g-kv where k is constnat

Derive a formula for the velocity of the body as a function of time assuming it starts from rest (v=0 at t=0)

v=?

- calc -
**drwls**, Tuesday, September 8, 2009 at 3:01am
Integral of dv/(g - kv) = integral of dt

Integrate both sides, from time from 0 to t; and v from 0 to v, for an equation for t in terms of v. Then invert the equation for v(t)

- calc -
**Joe**, Tuesday, September 8, 2009 at 3:29am
It would be great to see how this is done. The prof's way of doing it is way confusing and intricate and I just started learning integrals.

- calc -
**MathMate**, Tuesday, September 8, 2009 at 8:08am
This is why you'd need to do exercises in integration.

To give you a hint,

∫dv/(g-kv) = -log(g-kv)/k

This could be inferred from standard integrals:

∫dx/(a+bx) = (1/b)log(a+bx)

- calc -
**bobpursley**, Tuesday, September 8, 2009 at 9:15am
I agree. I suspect the Prof's method is confusing because you don't understand it....you learn by experience spaced over time. There is no substitute in calculus for hump, grunt, and strain...that is, practice, trial and error.

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