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...?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Ant crawls on a meter strick with acceleration

**Physics Forums | Science Articles, Homework Help, Discussion**