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Ant crawls on a meter strick with acceleration

  1. Oct 10, 2004 #1
    I have a question ..an ant crawls on a meter strick with acceleration

    [tex]a(t)=t-1/2t^2[/tex]. After t seconds the ants intiial velocity is 2cm/s.

    The ants initial poistion is the 50cm mark. So ten [tex]\int_{a}^{b}f(t)dt[/tex]

    and [tex]v(t)=1/2t^2-1/6t^3+c[/tex] and because [tex]v(0)=2m/s[/tex]

    the equation is [tex]v(t)=1/2t^2-1/6t^3+2[/tex] and I did the same thing for

    the position function and came up with the final function for positon of

    [tex]s(t)=1/6t^3-1/24t^4+2t+50[/tex]....the problem is when asked what

    was the ants average velocity over the first 3 seconds of its journey. Using

    the [tex]\frac{1}{b-a}\int_{a}^{b}f(x)dx[/tex]

    theorem.......[tex]\frac{1}{3}\int_{0}^{3}t-\frac{1}{2}t^2dt[/tex] then I

    got [tex]\frac{1}{3}(\frac{9}{2}-\frac{27}{6})-\frac{1}{3}(0)[/tex]...Right

    here where the lower limit is [tex]0[/tex] I dicided that since

    [tex]v(0)=2[/tex] I would enter that value in for

    it....[tex]\frac{1}{3}(\frac{9}{2}-\frac{27}{6})-\frac{1}{3}(2)[/tex] but

    my answer came out to be [tex]-\frac{2}{3}[/tex] wich I wouldnt get if i

    averaged the regualr function with out using the theorem

    [tex]\frac{1}{b-a}\int_{a}^{b}f(x)dx[/tex]. Am I wrong?

    And the second qustion is the same question except.."over the first 6 seconds of its journy" so [tex][0,6][/tex]...the graph for this function goes down into the negatives...?
     
    Last edited: Oct 10, 2004
  2. jcsd
  3. Oct 10, 2004 #2

    Gokul43201

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    Why are you averaging the acceleration, instead of the velocity ?
     
  4. Oct 10, 2004 #3
    Hmm thats seems like a dumb mistake :uhh: I should be integrating the velocity
    ay?
    [tex]\frac{1}{b-a}\int_{a}^{b}f(x)dx[/tex]

    [tex]\frac{1}{3}\int_{0}^{3}\frac{1}{2}t^2-\frac{1}{6}t^3+2dt[/tex]


    [tex]\frac{1}{3}(\frac{57}{8})-\frac{1}{3}(0)[/tex] comes out to be

    [tex]2.375m/s[/tex]. Does anyone have any comment on that answer? I think its right from looking at the graph.
     
    Last edited: Oct 10, 2004
  5. Oct 10, 2004 #4

    HallsofIvy

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    Use the basic definition of "average velocity". Since you already have the distance function, how far did the ant crawl between t= 0 and t= 3? Now divide the distance by 3 seconds.
     
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