Question

Calculus Help
*medal*

Answer 1

Positive velocity means the particle is moving `forward`.
Negative velocity means the particle is moving `backwards`.

So what happens when the velocity is zero?
It's changing directions, ya? :)

Answer 2

I got the derivative, s'(t)=3t^2-18t+24
Then I set it equal to 0 to get x=2, 4
Since its in reverse 2<t<4
So you plug in 3 into the s(t) regular equation, correct?

Answer 3

You found your critical values for your position function.
Those are the places where the velocity is 0.
Those are TWO locations where the particle changes direction.

The wording of this question is poor.
It changes directions twice, not just at a single location.
I assume maybe they want the time when the particle is going backwards.
That's probably what they mean by reverse :P

Answer 4

But how do you know that it's in reverse from t=2 to t=4?
What if it's in reverse every time except then? :)

Answer 5

Maybe it's going forward from 2 to 4.

Answer 6

You should do a quick `sign chart` to make sure ^^

Answer 7

It would be 3 still.

Answer 8

thats why you plug in 3

Answer 9

Plug in test points, make sure it's going in reverse between 2 and 4.

Answer 10

I don't quite understand why you're using this t=3 number.
It's either starting going in reverse at t=2, or at t=4.
t=3 shouldn't have anything to do with that.

Answer 11

2<t<4

Answer 12

So t would be 3

Answer 13

no no.
It says find the acceleration and position `at the moment` when it changes direction.

Answer 14

Not one second later, exactly at the moment it changes directions.

Answer 15

ohh. so i would plug in both 2 and 4 into the equation

Answer 16

We'll just skip past the sign chart I guess :)
If you plugged in test points, it should give you something like this:So yes, you are correct.
It starting going in reverse at t=2.

So we want to know the position and acceleration at t=2.

Answer 17

s(2)=?
Remember how to get your acceleration function after that?

Answer 18

Right. Which then give you 21m, correct?

Answer 19

For position? Good.

Answer 20

Right. Then Acceleration = 6t-18
Plug in 2 and you get -6 m/sec^2 right?

Answer 21

yay good job \c:/

Answer 22

actually they didn't give us a unit of measure,
so maybe don't use meters.

units / sec^2

Answer 23

So for the position it would be 21 units/sec^2

Answer 24

oh oh they didn't tell us how time is measured either haha,
so we can't assume it's seconds.

Ya I would just leave the units off altogether :)

no, position is measured in units, not units/sec^2.

Answer 25

It says to have units of measurement in my answer