Math Help - Velocity problem -- i think my answer is wrong

1. Velocity problem -- i think my answer is wrong

Here is the problem:

A body of mass m is moving with velocity v in a gravity-free laboratory. It is known that the body experiences resistance in its flight proportional to the square root of its velocity. Consequently, the motion of the body is governed by the inital-value problem:

$
dv/dt = -k*sqrt(v)
$

$V(0) = Vo$

where k is a positive constant. Find v(t). Does the body ultimately come to a rest? If so, when?

Okay...after separating and massaging, I got

$v(t) = ((kt +/- c * sqrt(Vo))^2)/4$ "k times t, plus or minus (c times the square root of V not)...all squared, divided by four"

and I took the limit of that to see if it would come to a rest. The limit is undefined, so I concluded that 'no' it will not stop.

This seems wrong since there IS resistance and the whole "if so..." part.

Plus, when I try to plug my equation back into the original problem to check myself, it doesn't quite work out as I would expect.

to find v(t), i did $dv/sqrt(v) = -kdt$, then I integrated both sides and solved for v to get: $v= ((-kt-c)^2)/4$. Then I used V(0) = Vo to solve for C and thats how I got my v(t).

Please let me know if you think I am right or wrong.

Thanks!

2. $\frac{dv}{dt}=-k\sqrt{v}$ implies that we can rewrite the equation as follows

$\int\frac{dv}{\sqrt{v}}=-k\int dt=-kt+c$
so after integrating we find

$2v^{\frac{1}{2}}=c-kt$
this we can then rewrite as

$v=\frac{1}{4}\left(c-kt\right)^{2}$

we can now compute $v(0)=\frac{c^2}{4}=v_{0}$

and we can clearly see that when $kt=c$ the velocity will be zero, at time $t=\frac{c}{k}$ the velocity will be zero.

3. Originally Posted by Mauritzvdworm
$\frac{dv}{dt}=-k\sqrt{v}$ implies that we can rewrite the equation as follows

$\int\frac{dv}{\sqrt{v}}=-k\int dt=-kt+c$
so after integrating we find

$2v^{\frac{1}{2}}=c-kt$
this we can then rewrite as

$v=\frac{1}{4}\left(c-kt\right)^{2}$

we can now compute $v(0)=\frac{c^2}{4}=v_{0}$

and we can clearly see that when $kt=c$ the velocity will be zero, at time $t=\frac{c}{k}$ the velocity will be zero.
Aha!

I see...I was making it harder than it needed to be.

Thank you very much