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

AI Thread Summary
The discussion centers on finding the coefficient 'a' in the drag force equation F=kv^a that allows a body to travel an infinite distance. Initial calculations suggest that if a=1, the velocity approaches zero too quickly, preventing infinite travel. Participants clarify that while a=1 leads to a bounded displacement, other values like a=2 also do not yield infinite distance. The key takeaway is that the relationship between velocity and time must be examined to determine the appropriate value of 'a' for infinite travel. Ultimately, the consensus is that a must be less than 1 for the body to achieve infinite distance under the influence of drag.
Patrikp
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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]
 
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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.
 
geoffrey159 said:
Hello,
Unless I'm wrong, if ##a=1## then ## v(t) = ue^{-\frac{k}{m}t}##

How did you get this?
 
Patrikp said:
How did you get this?
Ok I see why a=1 doesn't work now...
 
Patrikp said:
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.
geoffrey159 said:
##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.
 
haruspex said:
That's not a correct deduction.
Actually, it is a correct deduction.

Chet
 
Chestermiller said:
Actually, it is a correct deduction.

Chet
It may be a correct statement, but it is not correct as a deduction.
 
haruspex said:
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.
 
  • #10
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.
 
  • #11
| 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 = - kva o_O

and then a=2 is no good either !
 
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