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total distance is integral of velocity but u gotta take into consideration which way it's going

so you found where it changes direction right?

the thing is...|dw:1357322365854:dw|

if u take just the integral, the area on the bottom gets subtracted

i thought to find the distance u have to take integral of the velocity.

okay listen, where did the particle change direction?

rad(2pi) and 0

0?

yeah those were the solutions but only rad(2pi) works

no it was just t = 2.507

yeah that's rad(2pi)

so integrate that function from 0 to 2.507, then from 2.507 to 3

that's all he's doing by integrating the absolute value of V over 0 < t < 3

how'd u integrate the function tho?
like algebraically?

isn't that the calculator part?

i dont think u can integrate exponents inside trig functions yet.. lol

or do u just plug it into the graphing calculator?
ohhh -_-
ok.

i mean i dont think there is a way that anyone has found how to do that

lol yea

ohhh
yeah i got the answer when i plugged it into the calculator

You mean to tell me.... I WASTED MY TIME???!!!!!!

nooooooo......
like after you told me all that...

k good lol

sorry..if anyone's still viewing this
for part d)
why did they add 1 to rad(2pi) ?