An object moves through a medium with air resistance proportional to velocity. A $t=0$ the power is cut off so the only force acting on the object is air resistance. At $t=0$ the velocity of the object is $50m/s$ and at 10 seconds the object has slown down to $40m/s$
In the book we are given the formula:
$$\frac{dv}{dt}= \frac{-k}{m} – g$$
where k is a constant and g is the force of gravity $32 ft/s$
write a differential equation for the situation. I can't seem to figure out what the problem is Im not given k and im not given m so how am I supposed to do this? Am I using the wrong formula?
Best Answer
The problem has nothing to do with the formula.
Given that the only force acting the object is air resistance which is proportional to its velocity, you simply have
$$\frac{dv}{dt}=-av$$
Integrate the ode to get,
$$\ln v(t)=-at+C$$
Use the initial condition $v(0) = 50$ to obtain the constant $C$,
$$C=\ln 50$$
Then, use the information at $t=10$, i.e. $v(10) = 40$ to obtain the equation below for resistance constant $a$,
$$\ln 40 = -10a + \ln 50$$
which yields,
$$ a = \frac{\ln 5 - \ln 4}{10}$$
Therefore, the differential equation for the situation is just,
$$\frac{dv}{dt}= \frac{\ln (4/5)}{10}v$$