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anonymous
 5 years ago
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
anonymous
 5 years ago
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

This Question is Closed

Stacey
 5 years ago
Best ResponseYou've already chosen the best response.0Since s(t) is the position, we can take the derivative to get the equation for the velocity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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<t<3 or t>5, backward 3<t<5 (B) speeding up t>4, slowing down 0<t<4

Stacey
 5 years ago
Best ResponseYou've already chosen the best response.0Did you use interval notation? forward (0,3) union (5,infinity); backward (3,5)

Stacey
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, I need to correct that.

Stacey
 5 years ago
Best ResponseYou've already chosen the best response.0Forward is [0,3) union (5,infinity)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i think i used the wrong bracket

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what would speeding up and slowing down be then?

Stacey
 5 years ago
Best ResponseYou've already chosen the best response.0slowing down is [0,4) and speeding up is (4,infinity)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it says the sppeding up and slowing down ones are wrong :/ but the other two are correct

Stacey
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0maybe the square bracket shouldnt be used?

Stacey
 5 years ago
Best ResponseYou've already chosen the best response.0a(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
 5 years ago
Best ResponseYou've already chosen the best response.0yes 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 !
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