# Drag force F=kv^a

## 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 im not sure but i think that a=1 is good.
Is this ok?
[/B]

Last edited:

## Answers and Replies

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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.

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

How did you get this?
Ok I see why a=1 doesn't work now...

haruspex
Homework Helper
Gold Member
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
Mentor
That's not a correct deduction.
Actually, it is a correct deduction.

Chet

haruspex
Homework Helper
Gold Member
Actually, it is a correct deduction.

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

Chestermiller
Mentor
It may be a correct statement, but it is not correct as a deduction.
Right. They would have to show it mathematically.

Chet

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.

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