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)
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=?
4 answers
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.
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)
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)
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.