How long does it take for the speed of the train to decrease to $1~\text{m/s}$?

Question

Here's the full problem:

The speed of a train coasting on a level track satisfies the differential equation $\mathrm{d}V/\mathrm{d}t = kv^2$. The initial speed of the train is $v(0)=10~\text{m/s}$ and the speed is decreasing at the rate of $-1~\text{m/s}^2$ when $v=5~\text{m/s}$. How long does it take for the speed of the train to decrease to $1~\text{m/s}$? Under this model, when does the train stop?

I'm very bad with word problems and have been struggling quite a bit. I'd go to my school's tutoring center but currently we are on labor day vacation. -_-

EDIT 1: I've attempted it a few times and here's my answer: $vknot=10~\text{m/s}, v=5~\text{m/s}, a=-1~\text{m/s}^2$

I found $k = -1/25$ and the proceeded to integrate the $\mathrm{d}V/\mathrm{d}t = kv^2$. The equation I then got was $v=kv^2t + vknot$ and plugged all my conditions in. My answer was $t=5~\text{s}$. I have no idea if I am correct as word problems always throw me off.

EDIT 2: I found k=-1/25 by plugging into the dV/dt equation. dV/dt is the same as acceleration and it's stated that our acceleration is -1m/s^2, leading to this: $-1 = k(5)^2$ ===> $-1 = 25k$ ===> $-1/25 = k$. I don't know if this was valid to do, but I wasn't sure how to approach so made the assumption that it was valid.

EDIT 3: After realizing the error with my integration, after integrating I got the formula $v=v(vknot)kt + vknot$. Time was then t=2.5 after plugging in my information. Now I just don't know how to tell when the train stops...

Answer 1

Ok, I'll help with the integration, since that's obviously your weak point.

You have:

$\frac{dv}{dt}=k v^2$

Where $v\equiv v(t)$, so to solve the above differential equation, you need to separate the variables - all the $v$ parts on one side, and all the $t$ parts on the other

$\frac{dv}{v^2}=k dt$

This was probably the confusing part (I think that was confusing for me too when I was in my 1st year of Uni and when I had no clue to what DE were, so don't bring yourself down : ) )

Now you can safely integrate both sides. We'll set $v(0)\equiv v_0$ for shorthand notation:

$\int\limits_{v_0}^{v}\frac{dv'}{v\ '^2}=k\int\limits_{0}^{t}dt'$

After integration you have:

$-\left.\frac{1}{v'}\right|_{v_0}^v=k t'\left.\right|_0^t$

That is after you put limits in:

$-\frac{1}{v}+\frac{1}{v_0}=kt\Rightarrow v(t)=\frac{10}{1-10 kt}$

Now you can put $k$ in and try to graph it, and solve your problem :)