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Since s(t) is the position, we can take the derivative to get the equation for the velocity.

i did that and got this but the answer is not correct:
s(t)=2t 3 −24t 2 +90t
v(t)=s ′ (t)=6t 2 −48t+90=6(t−3)(t−5)
a(t)=s ′′ (t)=12t−48=12(t−4)
(A) forward 05, backward 34, slowing down 0

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Did you use interval notation? forward (0,3) union (5,infinity); backward (3,5)

Actually, I need to correct that.

Forward is [0,3) union (5,infinity)

oh i think i used the wrong bracket

what would speeding up and slowing down be then?

slowing down is [0,4) and speeding up is (4,infinity)

it says the sppeding up and slowing down ones are wrong :/ but the other two are correct

speeding*

maybe the square bracket shouldnt be used?

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