How long did it take to travel the distance? 
A particul moves on a straight line, the only force acting on it being a resistance proportional to the velocity. If it started with a velocity of $1,000$ ft. per sec. and had a velocity of $900$ ft. per sec. when it had travelled $1,200$ ft., calculate to the nearest hundredth of a second the time it took to travel this distance.

Attempt:
From the given we have $v(t) = 1000-v(t)k$ and thus $d(t) = \dfrac{1000t}{1+k}$. Then solving when $d(t) = 1200$ and $v(t) = 900$ gives us $t = \dfrac{4}{3} = 1.33$.
The correct answer is $1.26$ but I don't see what I did wrong in my solution.
 A: The equation you want to solve is
$$\ddot{x}(t) = -k\dot{x}(t),$$
or
$$\dot{v}(t) = -kv(t)$$
a damping force proportional to the velocity.
With the anzatz $v(t) = Ae^{at}$ we get that $a = -k$.
Then at $t=0$, $v=1000$ so $A = 1000$, or
$$v(t) = 1000e^{-kt}.$$
Integrating we get
$$x(t_1) - x(t_0) = -\frac{1000}{k}\left(e^{-kt_1} - e^{-kt_0}\right).$$
If we take $x(0) = 0$ then this becomes
$$x(t) = -\frac{1000}{k}(e^{-kt}-1).$$
This can be rewritten as
$$x(t) = \frac{1}{k}(1000 - v(t)).$$
Substituting $v(t) = 900, x(t) = 1200$ gives $k = 1/12.$
Finally, since the velocity is $900$ at this point,
$$900 = 1000e^{-t/12} \to -12 \cdot \ln 0.9 = t \approx 1.26.$$
A: We have:
$$
\frac{dv}{dt}=-kv \quad \rightarrow \quad v(t)=v_0e^{-kt}
$$
and, from the initial condition $v_0=1000$
$$
v(t)= 1000e^{-kt}
$$
also we have:
$$
\frac{dv}{dr}=\frac{dv}{dt}\frac{dt}{dr}=-kv\frac{1}{v}=-k \quad \rightarrow \quad v(r)=-kr+v_0=1000-kr
$$
so:
$$
v(1200)=900=1000-k1200 \quad \rightarrow \quad k=\frac{1}{12}
$$
and, from $v(t)$:
$$
900=1000 e^{-\frac{t}{12}} \quad \rightarrow \quad t=12 \ln(0.9)=1.264326...
$$
