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Velocity as a function of position problem (simple i think)

  1. Apr 6, 2013 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    If we know that velocity [itex]v[/itex] of a particle as a funciton of position [itex]x[/itex] is: $$v(x)=20-2/3x$$
    then i am asked to determine the acceleration when [itex]x=15[/itex]. I am then asked to show that the particle never actually reaches [itex]x=30[/itex]


    2. Relevant equations
    $$vdv=adx$$
    $$vdt=dx$$
    $$adt=dv$$

    and [itex]a[/itex] is the acceleration (not necessarily constant) and [itex]t[/itex] is time.


    3. The attempt at a solution
    I have already reduced the problem to what we see in [itex]1.[/itex] I have tried to work with the relevant equations but i cant make sense of the second two since they are time-dependent, which is a variable i have no information on. However, intuitively, i feel to answer the second part i would take a limit as t approaches infinity, so I am conflicted.

    Thanks everyone for the help! you people are awesome!!
     
  2. jcsd
  3. Apr 6, 2013 #2

    SammyS

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    v(x)=20-2/3x$
    By the way, is that
    [itex]\displaystyle \ v(x)=20-\frac{2}{3x}\ ?[/itex]​
    ... or is that
    [itex]\displaystyle \ v(x)=20-\frac{2}{3}x\ ?[/itex]​

    You need to find the velocity, v, when x = 15 ?


    Then, take the derivative of v with respect to t . (Both sides of the equation.)
     
  4. Apr 6, 2013 #3

    joshmccraney

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    i thought order of operations made it linear, that is,
    [itex]\displaystyle \ v(x)=20-\frac{2}{3}x\ [/itex]​
    but i guess youre probably used to posts getting sloppy?


    wait, no, i need to find acceleration when x = 15. i think that's what i posted

    yes, this i know, but there are no t's. i assume position is a function of time, and thus i have to reckon with the chain rule. can you elaborate on what you mean?
     
  5. Apr 6, 2013 #4

    SammyS

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    Yes. That's it exactly.

    If you work this out first;
    [itex]\displaystyle \frac{d}{dt}v=\frac{d}{dt}\left(20-\frac{2}{3}\,x\right)\,,[/itex]

    then you will see that you also need: the velocity, v, when x = 15 ?
     
  6. Apr 6, 2013 #5

    joshmccraney

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    ahhh yes that makes perfect sense (although i would have never thought of that). the answer is -20/3---thanks! but do you have any recommendations for showing that x = 30 is impossible? i was thinking of:

    [itex]\displaystyle \ 30=\int_0^\infty (20-\frac{2}{3}x)dt\ [/itex]​
    it seems if this is true we would have shown that the only time we will reach x = 30 is in the limit (i can finesse this at the end). any suggestions as to how i proceed?

    thanks so much for the advice!
     
  7. Apr 6, 2013 #6

    Curious3141

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    [itex]\displaystyle \ 30=\int_0^\infty (20-\frac{2}{3}x)dt\ [/itex]​

    How would you go about showing that?

    I would just solve the differential equation ##\displaystyle \frac{dx}{dt} = 20 - \frac{2}{3}x## to get ##x## in terms of ##t## then prove that ##x<30## for all ##t##.
     
  8. Apr 6, 2013 #7

    joshmccraney

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    you guys are too smart. thanks!
     
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