A particle moves along a straight line and its position at time t is given by

s(t)=2t^3−24t^2+90t t=0 where s is measured in feet and t in seconds.
(B) Use interval notation to indicate the time interval(s) when the particle is speeding up and slowing down.

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward
4 days ago

Question

A particle moves along a straight line and its position at time t is given by

s(t)=2t^3−24t^2+90t t>=0 where s is measured in feet and t in seconds. 
(B) Use interval notation to indicate the time interval(s) when the particle is speeding up and slowing down.

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward
4 days ago

stacey · Answer

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

anonymous · Answer

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

stacey · Answer

Your work is good.

stacey · Answer

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

stacey · Answer

Actually, I need to correct that.

stacey · Answer

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

anonymous · Answer

oh i think i used the wrong bracket

anonymous · Answer

what would speeding up and slowing down be then?

stacey · Answer

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

anonymous · Answer

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

anonymous · Answer

speeding*

stacey · Answer

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.

anonymous · Answer

maybe the square bracket shouldnt be used?

stacey · Answer

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.

anonymous · Answer

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 !