Ant crawls on a meter strick with acceleration

[Alem2000]

I have a question ..an ant crawls on a meter strick with acceleration

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

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

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

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

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

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

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

the $$\frac{1}{b-a}\int_{a}^{b}f(x)dx$$

theorem...$$\frac{1}{3}\int_{0}^{3}t-\frac{1}{2}t^2dt$$ then I

got $$\frac{1}{3}(\frac{9}{2}-\frac{27}{6})-\frac{1}{3}(0)$$...Right

here where the lower limit is $$0$$ I dicided that since

$$v(0)=2$$ I would enter that value in for

it...$$\frac{1}{3}(\frac{9}{2}-\frac{27}{6})-\frac{1}{3}(2)$$ but

my answer came out to be $$-\frac{2}{3}$$ which I wouldn't get if i

averaged the regualr function without using the theorem

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

And the second qustion is the same question except.."over the first 6 seconds of its journy" so $$[0,6]$$...the graph for this function goes down into the negatives...?

 

[Gokul43201]

Why are you averaging the acceleration, instead of the velocity ?

 

[Alem2000]

Hmm that's seems like a dumb mistake I should be integrating the velocity
ay?
$$\frac{1}{b-a}\int_{a}^{b}f(x)dx$$

$$\frac{1}{3}\int_{0}^{3}\frac{1}{2}t^2-\frac{1}{6}t^3+2dt$$


$$\frac{1}{3}(\frac{57}{8})-\frac{1}{3}(0)$$ comes out to be

$$2.375m/s$$. Does anyone have any comment on that answer? I think its right from looking at the graph.

 

[HallsofIvy]

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.