Question

Air Resistance acting on a falling body can be taken into account with the following relation to acceleration a=dv/dt=g-kv. Derive a formula for the velocity as a function of time

Answer 1

\[\int\limits_{0}^{v}dv=\int\limits_{0}^{t_2}a dt \] let a=u=g-kv. Setting g-kv=0 can lead to v=g/k so substituting the following expression for velocity in u gives \[\int\limits_{0}^{v}dv=\int\limits_{0}^{t_2}(g-k(g/k))dt\] I'm thinking of integration by parts \[(g-k(g/k))t -\int\limits_{0}^{t_2}t*d/dt(g-k(g/k))\] [g and k are both constants]

Answer 2

I'm specifically stuck with the integrand \[v*du\] with u=(g-k(g/k))