Find k & a for Infinite Distance Travel w/ Drag Force F=kv^a

[Patrikp]

Homework Statement

Only force acting on body is drag force $$F=kv^a$$ where k is constant, a coefficient bigger than zero and v is velocity. Body is given initial velocity u.
Find a such that body travels infinitely large distance.[/B]

Homework Equations

$$F=kv^a$$
$$F=m(acceleration)$$
$$acceleration=dv/dt=d^2s/dt^2$$[/B]

The Attempt at a Solution

Since distance is infinite I think that v should never reach zero.
From $$-m \frac{dv}{dt}=kv^a$$ I found v(t) :
$$v=(u^{-a+1}- \frac{kt}{m}(1-a))^{\frac{1}{-a+1}}$$
Here I looked at cases a<1,a=1,a>1 and I am not sure but i think that a=1 is good.
Is this ok?
[/B]

 

[geoffrey159]

Hello,
Unless I'm wrong, if $a=1$ then $v(t) = ue^{-\frac{k}{m}t}$ and it tends to 0 rather quickly, so it won't travel infinitely large distance.

 

[Patrikp]

Hello,
Unless I'm wrong, if $a=1$ then $v(t) = ue^{-\frac{k}{m}t}$

How did you get this?

 

[Patrikp]

How did you get this?

Ok I see why a=1 doesn't work now...

 

[haruspex]

Ok I see why a=1 doesn't work now...

Good, but you're still not at the answer. How v behaves with time is useful, but you need to get to how distance depends on time.

$v(t) = ue^{-\frac{k}{m}t}$ and it tends to 0 rather quickly, so it won't travel infinitely large distance.

That's not a correct deduction.

 

[Chestermiller]

That's not a correct deduction.

Actually, it is a correct deduction.

Chet

 

[haruspex]

Actually, it is a correct deduction.

Chet

It may be a correct statement, but it is not correct as a deduction.

 

[Chestermiller]

It may be a correct statement, but it is not correct as a deduction.

Right. They would have to show it mathematically.

Chet

 

[geoffrey159]

I understand Haruspex's remark: if for example the speed was $v(t) =\frac{1}{t}$, it would still tend to 0 as $t$ grows to infinity, but the displacement would be infinite. It doesn't prove anything.

The correct explanation would be that the displacement is :

$x(t) - x(0) = \int_0^t v(s) ds = - \frac{mu}{k}(e^{-\frac{k}{m}t} - 1)$.

Since $|x(t)-x(0)| \le \frac{mu}{k}$, the displacement is bounded independently of time, so $a$ cannot be equal to 1.

 

[geoffrey159]

Hello,

Try $a=2$ ! In that case, your differential equation can be written more simply: $\frac{k}{m} = \frac{d}{dt}(\frac{1}{v})$;
and the displacement will grow infinitely large as time grows.

 

[BvU]

| am somewhat surprised by the sign of k. If there is mention of drag force, I would expect the force vector to oppose the velocity vector. It is fine to use magnitudes, but then I would still expect to see F = - kv^a

and then a=2 is no good either !