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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 0
5, backward 3 4, slowing down 0
Your work is good.
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
I am not sure why it would say those are wrong. The second derivative is a line, so we just have to see where it is positive and negative. a(4) =0, which means we use a parenthesis next to the 4.
maybe the square bracket shouldnt be used?
a(0)=-48, so we are clearly slowing down at t=0. Also we can look at the graph of the velocity and that is a parabola. Looking at that graph, we can also see that the velocity is decreasing (slowing down) on the interval [0,4) and increasing (speeding up) on the interval (4,infinity). It sounds like you are on a computerized homework. Sometimes those are wrong. I would double check what it is asking for and what I'm typing. Otherwise, it may just be wrong.
yes i am on a computerized homework and maybe I will check to see if it's maybe it's worng instead of me but thanks anyway !