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.