The acceleration of an object of mass M kg is partially impeded by air resistance proportional to the velocity$\space v \space m/s$ of the object. The motion is modelled by the differential equation $\frac{dv}{dt} = g - \frac{k}{M}v$, where M, g and K are constants.
b) If an object of mass $4kg$ falls from rest with air resistance that is $2$% of its velocity, determine the time when its velocity is $25m/s$. Assume that $g=9.8$
So first I determine the expression for velocity in terms of time by integrating using separation of variables. $v = -\frac{MA}{\:k}e^t+\frac{Mg}{k}$, where $A = e^c$ which results in 2 unknown variables $A$ and $k$
So my question is how can I use the information: "$2$% of its velocity"? Is the air resistance = $\frac{k}{m}$?
BTW my textbook gives $v=\frac{Mg}{k}-Ae^{-\frac{k}{M}t}$ and the answer is $8.12$
I would check your differential equation
$$\frac{dv}{dt} = g - \frac{k}{m}v$$
$$\implies \int\frac{dv}{mg - kv} = \int\frac{dt}{m}$$
$$-\frac{1}{k}\ln (mg-kv) = \frac{t}{m} + C$$
$$\implies mg-kv = Ae^{-\frac{kt}{m}}$$
$$v(t) = \frac{mg}{k} - \frac{A}{k}e^{-\frac{kt}{m}}$$
Let $\frac{k}{m} = D$
$$\implies v(t) = \frac{g}{D} - \frac{A}{mD}e^{-Dt}$$