Question

please help explain a calc question!!

Answer 1

@ganeshie8 @myininaya @SolomonZelman can you please explain this?! nobody has been able to explain it correctly!

Answer 2

@SithsAndGiggles please explain!!

Answer 3

@freckles please please explain!

Answer 4

\[\text{ average velocity on the interval from t=a to t=b is } \\ \frac{s(b)-s(a)}{b-a}\]

Answer 5

\[\text{ instantaneous velocity at t=c is } \\ \frac{ds}{dt} |_{t=c} \]

Answer 6

\[\text{ also we know } \frac{ds}{dt}=v\]

Answer 7

\[ p \text{ is moving to left when } s \text{ is decreasing } \\ p \text{ is moving to right when } s \text{ is increasing }\]

Answer 8

@ganeshie8 help!!!

Answer 9

@Michele_Laino @ganeshie8

Answer 10

Please you have to perform the integration of the function v(t), in order to find s(t), first

Answer 11

what does that mean

Answer 12

what is the result of this integral, please?
\[\int\limits v(t) \quad dt=\int\limits (t ^{2}-9t+18)dt\]

Answer 13

we are doing ONLY NUMBER 3 correct? @Michele_Laino

Answer 14

ok! Integration is correct!

Answer 15

that's your s(t), now in order to know the average speed, you can apply this formula:
\[average \quad speed =\frac{ s(8)-s(0) }{ 8-0 }\]

Answer 16

what is s?

Answer 17

s is the name of the function that you have found after integration, namely:
\[s(t)=\frac{ t ^{3} }{ 3 }-9\frac{ t ^{2} }{ 2 }+18t\]

Answer 18

i dont get how to apply that to my equation

Answer 19

please set t=8 in order to find s(8), and set t=0, in order to find s(0)

Answer 20

0/8?

Answer 21

@Michele_Laino

Answer 22

please what is s(8)=?

Answer 23

10/3 i already did this for question 1

Answer 24

NUMBER 3

Answer 25

\[s(8)=\frac{ 8^{3} }{ 3 }-9\frac{ 8^{2} }{ 2 }+18*8=...\]

Answer 26

do you know what youre doing?

Answer 27

THE TIME PARTICLE WHEN THE PARTICLE IS MOVING RIGHT @Michele_Laino

Answer 28

please when s(t) is positive?

Answer 29

10/3

Answer 30

namely you have to solve this inequality:
\[s(t)>0\]

Answer 31

that is right

Answer 32

x=6 and x=3

Answer 33

@Michele_Laino hello

Answer 34

please solve this inequality:
\[t(2t ^{2}-27t+108)>0\]

Answer 35

2t^3-27t^2+108t > 0

Answer 36

please you have to find the interval or intervals of t for which s(t) is positive

Answer 37

how

Answer 38

please solve this equation:
\[t ^{2}-27t+108=0\]
first

Answer 39

solve for what

Answer 40

oops...solve the equation below:
\[2t ^{2}-27t+108=0\]

Answer 41

what is its discriminant?

Answer 42

-135

Answer 43

nevermind ill ask someone else, youre probably the worst xplainer on here lol

Answer 44

ok! then we have not zeros for that equation, namely our quadratic polyomial is neve equals to zero,it is alwys positive for all real values of t. so the iequality s(t)>0 is checked if and ony if t>0

Answer 45

ok so now what

Answer 46

your answer is the interval [0, +infinity)

Answer 47

... i dont think so ill ask someone else

Answer 48

please note that s(t)>0 if and onlyif t>0

Answer 49

please note that the function s(t) is:
\[s(t)=\frac{ t ^{3} }{ 3 }-\frac{ 9 }{ 2 }t ^{2}+108t+1\]
since s(0)=1, as required from the text of the problem, so in order to answer to point c), you have to solve this inequality:
\[\frac{ t ^{3} }{ 3 }-\frac{ 9 }{ 2 }t ^{2}+108t+1>0\]

Answer 50

copying and pasting what I have above:

\[p \text{ is moving to left when } s \text{ is decreasing } \\ p \text{ is moving to right when } s \text{ is increasing }\]

that p is the particle
s is decreasing when v<0
s is increasing when v>0


by the way v is another way to say s'

I don't remember what you have for v...
but pretend v=t^2+5t+6

first thing to do is find when v=0
0=t^2+5t+6
0=(t+3)(t+2)
So v=0 when t=-3 or t=-2 .

This means at t=-3 or t=-2 s is moving no where since the velocity (v) =0
Now we test around these numbers to see if v>0 or if v<0
now remember v>0 implies s increasing
and v<0 implies s decreasing



Draw a number line

------|-------|-------
-3 -2

We have three intervals to test
First interval: (-inf,-3)
Second interval: (-3,-2)
Last interval: (-2,inf)

So testing first interval.
Pull a number from (-inf,-3)
-4 is in that set of numbers
what is v(4)
well our v (velocity function) for this example is v=t^2+5t+6 for position function s=s(t).
v(4)=4^2+5(4)+6=16+20+6=36+6=42
v(4)=42.......42 is positive so that means s is increasing on the interval we labeled "First interval" aka (-inf,-3)


So now second interval.
Pull a number from (-3,-2)
-2.5 is a number in that set of numbers
what is v(-2.5)
v(-2.5)=(-2.5)^2+5(-2.5)+6=6.25-12.5+6=6.25-6.5=-0.25
v(-2.5)=-0.25 ..... -0.25 is negative so s is decreasing on the interval we labeled "Second interval" aka (-3,-2)


So now third interval.
Pull a number from (-2,inf)
0 is a number in that set of numbers
v(0)=(0)^2+5(0)+6=0+0+6=0+6=6
v(0)=6......6 is positive so s is increasing on the interval we labeled
"Last interval" aka (-2,inf)


----Summing it up (that is the answer for our example)
Since v was positive on (-inf,-3) U (-2,inf)
then s was increasing on (-inf,-3) U (-2,inf)

Since v was negative on (-3,-2)
then s was decreasing on (-3,-2)


---
We know this because v is the derivative of s
s=position function
s'=v=velocity function
didn't need this info but:
s''=a=acceleration function

Answer 51

oops i just saw that you responded to this question

Answer 52

@freckles i just posted it again because i deleted it before

Answer 53

wait im confused

Answer 54

can we do it together with my question? t^2-9t+18

Answer 55

Factor your v first
That is factor t^2-9t+18

Answer 56

Then you can do everything I did to find when v is greater than zero or just realize v is a parabola that is concave up
Which mean that everything not betwwen the zeroes is positive

Answer 57

i think peri answered it(: thank you