Question

I got an answer but i need help for interval.

A particle moves along a straight line and its position at time t is given by
s(t)=2t^3−30t^2+126

Answer 1

Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward.
Forward:
backward

Use interval notation to indicate the time interval(s) when the particle is speeding up and slowing down.
Speeding up:
slowing down:

Answer 2

\[s(x)`=6(t-3)(t-7)\]
and
s(t)``=\[12(t-5)\]

Answer 3

foward i got [0,3)(7,infiity) was right

Answer 4

@tkhunny please help?

Answer 5

Did you present the right equation? That factored first derivative has very little to do with the s(t) provided.

Answer 6

the equation was
\[2t^3-30t^2+126t\]
\[t \ge0\]

Answer 7

sorry about that

Answer 8

Okay, let's first talk about s"(t) - the second derivative

It's linear. You should be able to determine when this is negative and when this is positive. This should be simple. It better be!

Positive is speeding up
Negative is slowing down.

Answer 9

Oh I didnt know

Answer 10

so far I understand

Answer 11

For t > 5, s"(t) > 0 and we have positive acceleration (speeding up).
For t < 5, s"(t) < 0 and we have negative acceleration (slowing down).

Answer 12

ok, t>5 is counted as interval?

Answer 13

\((5,\infty)\)

For t = 5, nothing is going on with acceleration. s"(t) = 0

Answer 14

( 5 infy) for speeding up?

Answer 15

I already said that a couple of times. Why do you continue to doubt?

Answer 16

I entered it and it says it was wrong :/

Answer 17

That's a good answer. Maybe it wants \([5,\infty)\) and it considers s"(t) = 0 as some how non-negative acceleration.

I really don't like online things like this, We can talk and understand all we want, but figuring out what input is required turns out to be the hard part. Very annoying!

Answer 18

Yes , it is aboustely annoying. i tired [5,inf) but was also wrong :/

Answer 19

any ideas?

Answer 20

Nope. I can only get the right answer. We'll have to get the test designer to tell us what it wants. It could be wrong. Maybe we're still using the wrong equation.

Maybe it wants the whole thing at the same time?

Slowing down: \((0,5)\)
Speeding up: \([5,\infty)\)

Answer 21

The equation is right, but i entered it but i keep getting wrong

Answer 22

Good luck with that. Wave a magic wand at it, I guess. Did you check the air in the tires?

Really, I am sorry it isn't working. It's hard to learn math with silly stuff in your way.

Answer 23

its alright, thanks for your time.

Answer 24

Maybe it wants the Forward/Backward intervals first?

Answer 25

I got the answer for the forward one but not backward.

Answer 26

s'(t) = 6(t−3)(t−7)

Forward: \([0,3) \cup (7,\infty)\)
Backward: \((3,7)\)
Again, it's not perfectly obvious how it wants us to treat t = 3 and t = 7.

Answer 27

Thank you! the backward was right, as you said

Answer 28

the problem is just slowing down and speeding up

Answer 29

Oh, well, there goes that theory. Punt!

Answer 30

haha

Answer 31

are you familiar withLogarithmic Derivatives