s(t)= 3t^2+2t-2 a. find the velocity for time t b.find the acceleration for time t c. compute the total distance I derived the velocity equation and the acceleration as 5t+2 and 5 but the solution checker online says its wrong....
for c the interval is [0,1]
v=s'=3(2)t+2=6t+2
a. first derivative b. 2nd derivative c. s(1)
your number in front of t is a little off
wow.... this is what studying physics for 30 hours does to you....
how did i not see that rofl
you guys probably are facepalming right now. "how is this kid in calc 1??"
Hey can you do part c I think Im making another alegbra error and I cant see it I'm just not sure what to do with v=0 @ t=-1/3
s(1)-s(0)???????
as a thought jesus man you need more question marks?
Im pretty sure you can read more than just numbers as I said above I have been studying physics for 30 hours.
maybe take a break!
breaks are for sane people
s(t)= 3t^2+2t-2 v = s'(t) = 6t^2 + 2 a = v'(t) = 12t
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