Solve for velocity, if acceleration is a function of velocity? Yes, this is a canned question, because canned questions are simply solvable to understand ideas. No, this is not homework.
For simplicity, let's assign values.
A given object weights 1000 kg, it only has one force acting on it, and that's friction: $F_{t} = -200*v(t)$. 
Let's say, after 5 seconds, what is its final velocity? 
In case you haven't brushed up on your physics lately, here's a rundown of my mathematical problem.
F = ma, so a = F/m. With the numbers above, that means $a = \dfrac{-v}{5}$.
If I want to know final velocity after 5 seconds, it will basically be 
$$30 - \int_0^5 -v/5 \,dt= ??$$ ..two different variables.
I believe the answer should involve ln (or e), but I can't seem to set it up correctly. 
 A: Oh wow!
The answer just came to me!
I'm not sure if this will help anyone, but I figure it's probably better to just provide the answer for posterity.
So, really what we're talking about here is:
$$\frac{dv}{dt} = \frac{-v}{5}$$
So, $$\frac{dv}{v} = \frac{-1}{5}dt$$
Then integrate both sides to get
$$ln(v) = \frac{-1}{5}t + C$$
which rearranges to 
$$v = C*e^{-t/5}$$
And C can be solved with the initial condition of t = 0, v = 30. Which means C = 30. 
So, our equation for velocity at any time becomes 
$$v = 30e^{-t/5}$$
A: The acceleration of an object is the derivative of its velocity. The force acting on the object produces an acceleration according to the usual rule $F = ma$. The equation you get is
$$1000 v'(t) = -200 v(t).$$
In other words, $v'(t) = - \frac 15 v(t)$ which has the general solution $v(t) = v(0)e^{-\frac 15 t}$.
If $v(0) = 30$, you get $v(5) = \frac{30}{e}$.
A: Recall the definition of $a$:
$a = \dfrac{dv}{dt}; \tag{1}$
thus, with the numbers provided,
$\dfrac{dv}{dt} = -\dfrac{1}{5}v; \tag{2}$
this yields
$v(t) = v(t_0)e^{-(t - t_0) / 5}; \tag{3}$
here I have supplied the number not explicitly provided, which is $v(t_0)$, the initial velocity at time $t = t_0$; apparently $v(t_0) = 30 M/sec$, assuming we measure distance in meters (M). Plugging this and $t = t_0 + 5 sec$ into (3) yields
$v(t_0 + 5) = 30e^{-1}M/sec = \dfrac{30}{e}M/sec. \tag{4}$
The velocity decreases by a factor of $e$.
