Question

Please HELP! This is due in a couple of hours and I really need help. I have been trying to figure this out all day.
I am completely confused as to what to do on this someone please help!
A particle moves along a straight line and its position at time t is given by
s(t)=(2t^3)-(21t^2)+72t t>0 where s is measured in feet and t in seconds.
Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward.
And Use interval notation to indicate the time interval(s) when the particle is speeding up and slowing down.

Answer 1

looks like you need use some differentiation, finding the velocity and acceleration equations...

Answer 2

so for velocity it would be 6t^2-42t+72
and acceleration would be 12t-42
but how would I find the intervals for moving forward and back and speeding up and slowing down?

Answer 3

Realize that when velocity = 0, it may have changed direction. When acceleration = 0 it may have changed from accelerating to decelerating of vice versa.

Answer 4

our teacher tells us to use sign charts for the velocity and acceleration. where the signs are different, it's slowing down

Answer 5

I don't know I am really confused our teacher never showed us anything like this problem

Answer 6

if the velocity is positive, it's moving forward; if the velociuty is negaitve, it's moving backwards

Answer 7

for example for this one, figure out the critical points, which is where each of the equations for acceleration and velocity equal 0

Answer 8

do you know what to do next?

Answer 9

So for the velocity you would the critical points be 3 and 4?

Answer 10

yes

Answer 11

So then how would you know how to write the interval notation for this?

Answer 12

I am still way lost I don't know what to do with the critical points

Answer 13

HELP!

Answer 14

Have you found critical points yet?

Answer 15

Yes 3 and 4 for velocity and 3.5 for acceleration

Answer 16

At t = 3 -> S (3) = ...?
At t = 4 --> S (4) = ...?
Which one is max, min?

Answer 17

4 is the max and 3 is the min correct?

Answer 18

Did you find S(3) , S(4) ?

Answer 19

S(4)=-40
S(3)=81
so 3 is the max and 4 is the min

Answer 20

What is the next step?

Answer 21

Since t = 3.5 is inflection point and concavity downward, it confirm that ( 4, -40) is local minimum!

Answer 22

so is that then when it is going backwards?

Answer 23

The velocity will increasing in ( 0, 3) Max and decreasing from ( 3, 4) min point, then increasing again

Answer 24

So would you use these points to also show when it is speeding up and slowing down acceleration or do you have to involve the 3.5 critical point only?

Answer 25

Speeding up : t ε [ 0, 3 ] U ( 4, +inf ]
Slowing down: t ε ( 3, 4 ]

Answer 26

My homework for some reason is saying that it isn't speeding up on (0,3) U(4,infinity)

Answer 27

It doesn't tell me but could it be speeding up still when it is going backwards?

Answer 28

I think it is just when the values are negative then that is backward and when the values are positive it is forward and then acceleration would just be how fast it is going across the whole interval. I am not totally sure though?

Answer 29

ok

Answer 30

Got it it is speeding up on interval (3,3.5)U(4,infinity)
and it is slowing down on interval (0,3)U(3.5,4)

Answer 31

Great job :)

Answer 32

About the forward and backward, you can try forward to 81 and backward to -40!

Answer 33

or backward from the forward point -81- 40 = - 121