I have a question ..an ant crawls on a meter strick with acceleration(adsbygoogle = window.adsbygoogle || []).push({});

[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...?

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# Ant crawls on a meter strick with acceleration

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