Integration problem

  • Thread starter yoleven
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  • #1
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Homework Statement


A particle is moving along a straight line such that its acceleration is defined as a=(-2v)m/s^2.
If v=20 m/s when s=0 and t=0, determine the particle's velocity as a function of position and the distance the particle moves before it stops.

The answer is v=(20-2s)m/s ; s=10m

Homework Equations


a=[tex]\frac{dv}{dt}[/tex]

a=[tex]\frac{dv}{ds}[/tex]*[tex]\frac{ds}{dt}[/tex] which becomes...

ads=vdv


The Attempt at a Solution


given: a=(-2v) m/s
v=20 m/s
t=0
s=0
solution attempt:

a=[tex]\frac{dv}{dt}[/tex]

-2v*dt=dv

dt=-[tex]\frac{dv}{-2v}[/tex]

[tex]\int dt[/tex] = [tex]\int\frac{dv}{-2v}[/tex]


That's as far as I can get. If I evaluate the left hand side from 0 to t, I get t.
If I evaluate the left, I get messed up. If I pull out the -[tex]\frac{1}{2}[/tex],
I am left with [tex]\frac{1}{v}[/tex]. I think the integral of that is ln v.
I need some direction on this one please.
 

Answers and Replies

  • #2
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This problem is about damping. You have an object moving against a force proportional to its velocity, but directed opposite to it: F = -bv. The solution is an exponentially decaying one. This is what you got!

-2t = ln v + C ===> [tex]v = v_0 e^{-2t}[/tex]
 
  • #3
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I need velocity as a function of position.
How about this?
Can I use the equation v^2=v0^2+2a([tex]\Delta s)[/tex]

then I get v= [tex]\sqrt{200-4s}[/tex]

which would give me v=20-2s which is the answer in the book.

My concern here is the "s" and the value for "a" is -2v not just -2.
How do I deal with this under the square root sign?
 
  • #4
6
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You can't use that equation because that only applies when one has constant acceleration.

(1) [tex] v= v_0 e^{-t/\tau}[/tex]

(2) [tex] x = v_0\int_0^t e^{-\frac{t'}{\tau}} dt' = v_0\tau(1 - e^{-\frac{t}{\tau}})[/tex]

Solve for time in second equation (2). Then you can substitute that into (1). It will get hairy, but you should get an answer like the one in your book. Remember: [tex]e^{\ln{x}} = x[/tex].
 

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