Question

@ganeshie8

Answer 1

do you know how to find the velocity function
if y' =+ then particle is moving to the right (or find where y is increasing)
if y' =- then particle is moving to the left (or find where y is decreasing)

Answer 2

ignore the am

Answer 3

wow
didn't notice something...

Answer 4

there are x's and t's in that function :(

Answer 5

what??

Answer 6

yess, do you not know how to do it?

Answer 7

well this means either the teacher meant all t's instead of all x's
or it is a trick and x is a constant

Answer 8

i dont knoww :((

Answer 9

the question looks messed up, but maybe we can assume all t's just to keep going

Answer 10

i think @ganeshie8 is right
using our esp skills we should assume t's

Answer 11

anyways what did you get for s'?

Answer 12

where x=t

Answer 13

\[s(t)=-\frac{t^3}{3}+\frac{13}{2}t^2-30t\]

Answer 14

s'(t)=?

Answer 15

what is the derivative of t^3?

Answer 16

37t^2/2-30

Answer 17

\[s'(t)=\frac{-1}{3}(t^3)'+\frac{13}{2}(t^2)'-30(t)' \\ \]
t'=1 right
(t^2)'=?
(t^3)'=?

Answer 18

use power rule

Answer 19

(t^n)'=nt^(n-1)

Answer 20

nevermind im skipping it

Answer 21

why

Answer 22

it isn't that bad of a question

Answer 23

well if the assumption is right anyways

Answer 24

its so confusing

Answer 25

we want to find s'
then find where s' is increasing

Answer 26

you can do this

Answer 27

I'm right here and I will help you

Answer 28

my answer key is telling me it is A but i dont get how

Answer 29

It is definitely A
And I can lead you to that answer

Answer 30

but first you have to find
(t^2)' and (t^3)'

Answer 31

it's not that big of a deal because it is a pre test and i like learning it before but ill read the chapter through now and if i still dont understand it ill come back and tag you (:

Answer 32

\[s'(t)=\frac{-1}{3}(t^3)'+\frac{13}{2}(t^2)'-30(t)' \\ s'(t)=\frac{-1}{3}(3t^2)+\frac{13}{2}(2t)-30(1)\]
Ok well you can also look at this if you want to
you will see lot of the fractions cancel there

Answer 33

\[s'(t)=-1(t^2)+13(t)-30(1) \\ s'(t)=-t^2+13t-30\]

Answer 34

s'=0 when -t^2+13t-30=0
you can find the zeros of -t^2+13t-30 by solving that equation
then you should know our parabola is open down
the interval that has the function above the x-axis is our interval in which the particle moves right

Answer 35

Anyways you read the section and you can ask the question later again

Answer 36

im actually following! thanks for the explanation!