v(t) = t^3 - 3t^2 +2t 0 < t <3
Alright, I was fine finding the displacement ...
I did sigma(t^3-3t^2 + 2t)dt
I got t^3-3t^2 + 2t and evaluated on 0-3
(81/4 - 108/4 + 36/4) = 9/4 meters
However I am having trouble finding the total distance traveled....
Since it is over the period of 0-3, I know I need to split up the integrals when the sign of v(t) changes |(t^3-3t^2 + 2t)dt|
Setting v(t) to 0 I get 0, 1, 2 ... The sign changes from negative to positive at t=1
I took sigma 0,1 |t^3-3t^2 + 2t dt|- sigma 1,2 |t^3-3t^2 + 2t dt|
and got some funky answer
Not sure what my mistake was... please help...I have a test on this tomorrow and I am so lost
The solutions manual says the answer is 11/4 meters