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Drag force F=kv^a

  • Thread starter Patrikp
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  • #1
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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 im not sure but i think that a=1 is good.
Is this ok?
[/B]
 
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Answers and Replies

  • #2
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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.
 
  • #3
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Hello,
Unless I'm wrong, if ##a=1## then ## v(t) = ue^{-\frac{k}{m}t}##
How did you get this?
 
  • #4
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How did you get this?
Ok I see why a=1 doesn't work now...
 
  • #5
haruspex
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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.
 
  • #6
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That's not a correct deduction.
Actually, it is a correct deduction.

Chet
 
  • #7
haruspex
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Actually, it is a correct deduction.

Chet
It may be a correct statement, but it is not correct as a deduction.
 
  • #8
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It may be a correct statement, but it is not correct as a deduction.
Right. They would have to show it mathematically.

Chet
 
  • #9
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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
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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
BvU
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| 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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