Dynamics Problem

  • Thread starter ncr7
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  • #1
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Hey I have been looking at the problem for a long time and for some reason I can't think of how to solve for it... I know I have to do some Calc to figure it out.

Ok so here it is

The acceleration of a particle is defined by the relation a = -0.05[tex][v]^{2}[/tex], where
a is expressed in m/[tex]^{2}[/tex] and v in m/s. The particle starts at s=0m with a velocity of 5 m/s. Determine (a) the velocity v of the particle after it travels 10 m, (b) the distance s the particle will travel before its velocity drops to 2 m/s, (c) the distance s the particle will
travel before it comes to rest.

so far I think what I have to do is take the derivative of a which ends up being -.1v but after that I am stuck because now I have velocity as a function of velocity... any ideas? I realize I am most likely doing this totally wrong.
 

Answers and Replies

  • #2
Doc Al
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Since [itex]a = dv/dt[/itex], you'll need to integrate to find v. Start there.
 
  • #3
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so I'll get
[tex]\int[/tex]dv=[tex]\int[/tex]a dt
(v2-v1) = -0.05[tex]v^{2}[/tex]*t - 0 (since [tex]t_{0}[/tex]=0)

but I now have 2 unknowns though.
v and t

since I know the change in position shouldn't I do an integration over that? I'm just unsure how exactly it should be done
 
  • #4
Doc Al
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so I'll get
[tex]\int[/tex]dv=[tex]\int[/tex]a dt
(v2-v1) = -0.05[tex]v^{2}[/tex]*t - 0 (since [tex]t_{0}[/tex]=0)
:yuck:

Before you can integrate properly, you must separate the variables:
[tex] \frac{dv}{dt} = -0.05v^2[/tex]

[tex] \frac{dv}{v^2} = -0.05 dt[/tex]

Now try integrating. (And don't forget the integration constants.)
 
  • #5
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ok, I'm starting to have some wierd mistakes because I have been on this problem to long lol... but I don't have two integration constants for t or v. I have the change in s from 0m to 10m. This would only really be able to give me where it's time when it comes to a complete stop which is part of the problem. But right now I need to find velocity at 10m.
 
  • #6
Doc Al
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One step at a time. :wink: First complete the integration to find v as a function of t. Then, using v = dx/dt, integrate once more.
 
  • #7
Shooting Star
Homework Helper
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In this case, it would save a lot of trouble if he writes a = v(dv/dx). All the answers want some x at some v.
 
  • #8
Doc Al
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Even quicker! :smile: (I was trying to keep it as straightforward and simple as possible.)
 
  • #9
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ok I think I got it.

so taking

[tex] \int_0 ^{10m} -0.05v^2 dx = \int_0^v v dv [/tex]

I get


[tex]-0.05x|_0 ^{10} = ln|v||_{5m/s} ^v [/tex]

[tex] e^{1.11} = v [/tex]

getting 3.03 m/s

I think that's right...
 
Last edited:
  • #10
Doc Al
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Looks good, but be careful how your write up your work.
[tex] v\frac{dv}{dx} = -0.05v^2[/tex]

[tex] \frac{dv}{v} = -0.05dx[/tex]

[tex] ln(v) = -0.05x + C[/tex]
 
  • #11
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ok yeah I'm going to write it more throughly when I put it on paper. There wouldn't be the C because I have known points I'm integrating between. thanks for checking my work.
 

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