Question

A particle is moving with velocity v(t) = t^2 – 9t + 18 with distance, s measured in meters, left or right of zero, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t = 0 sec is 1 meter right of zero, that is, s(0) = 1.

Answer 1

The average velocity over the interval 0 to 8 seconds

The instantaneous velocity and speed at time 5 secs

The time interval(s) when the particle is moving right

The time interval(s) when the particle is
going faster
slowing down

Find the total distance the particle has traveled between 0 and 8 seconds

Answer 2

For the first one I'd take the integral of the function from 0 to 8 right?

Answer 3

@ganeshie8 Can you give this a look?

Answer 4

well the instantaneous velocity at t=5 can be found by pluggin in t=5 the velocity function is the one given...

Answer 5

also more math <.>

Does this help? ^-^ a(t) = 2t - 9 ---> a(0) = -9 b/ x(t) = t^3/3 - 9t^2/2 + 18t + c , x(0)=1 ---> c=1 average velocity = (x(8)-x(0))/8 = ... c/ v(5) = 5^2 - 9(5)+18 = ...

Answer 6

I see what you're doing but I don't think it's leading me to the right answer. You're saying the derivative of the velocity is equal to acceleration which is true. You also solved the integral but how does c=1? It is a definitive integral where C isn't included in the final answer no? I have a good amount of prior knowledge on the subject just need someone to point me to the right direction. Is there anything else you could share for the time intervals? That's the part that really makes me struggle.

Answer 7

And thanks for actually helping, after 2 hours you'd think there'd be more responses.

Answer 8

Yeah I'm sorry I'm Trying to explain slope.

Answer 9

First one should be 224
According to @MaydayPaRAYde the second part should be equal to -2

Answer 10

First on needs to be redone, just doesn't seem right to me.

Answer 11

one*

Answer 12

@IrishBoy123 do you have an idea?

Answer 13

this is lazy but at least it's the right way to go

without a plotter, you should look at intercepts etc and plot it manually

d is displacement

naturally, usual caveat -- check the working for yourself

then start answering......💥

Answer 14

https://www.desmos.com/calculator/hons6cwrqv

i meant "this"

Answer 15

Whoa..... hella confused now

Answer 16

Basically graph and solve?

Answer 17

Found my mistake for the first one. It'll be 26.66*1/8

Answer 18

#1 = 3.33
#2 = v= -2 s=2

Answer 19

graphing really has to be part of it. this is applied maths.

note however i did the integration as well

BTW x on the graph represents time. i got funny stuff from desmos when i used t. i don't know desmos that well to have a quick fix.

Answer 20

That's what I was planning to do, graph it to show the intervals and when it increase and decreases.

Answer 21

I looked up how to do it on the web! I tried but honestly I'm not to this yet but I wanted to help you. I'm sorry that's how it taught me I am strugling o understand... :( I'm sorry

Answer 22

It's cool, I am working through it. The points above the x-axis is where it is moving to the right, and the points below it is when it's too the left.

Answer 23

Oh... That's Helpful.

Answer 24

@IrishBoy123 what does the graph of the integral represent? displacement?

Answer 25

yes!!!


displacement!!

very important distinction

Answer 26

Ok so I would use that for the last part right?

Answer 27

Only problem now is the 4th one.

Answer 28

Ohhhhh

Answer 29

Derivative of velocity is equal to acceleration.

Answer 30

the distance is the area under the blue graph

https://www.desmos.com/calculator/el3uwmejba

there are a number of ways to get to it, i'm sure

i don't know if you are integration but where as
\(d(t) = \int dt \qquad v(t)\)
\(|d(t)| = \int dt \qquad |v(t)|\)

another reason why the plot is so useful.

Answer 31

sic
*integration*
integrating

Answer 32

soz

by 4th you meant "The time interval(s) when the particle is
going faster
slowing down "

yes use the graph or differentiate and solve

Answer 33

gtg for a while

you seem in command

Answer 34

Ight, just one more thing. Distance is always positive right?

Answer 35

Lel
Pretty sure it is.

Answer 36

yes, it is a scalar.

Answer 37

ie no direction, like the boy band.....😁

Answer 38

Oh god.....that was terrible....

Answer 39

Still having trouble or did you bump it on accident?

Answer 40

Bumped for fun. I got this down.

Answer 41

@IrishBoy123 Can't I just split the intervals then add?
Like from 0 to 3, 3 to 6, 6 to 8?

Answer 42

yup

Answer 43

That would give me distance correct?

Answer 44

yes

Answer 45

\[\text{Distance =} \int v + \int (-v) \text{[for when it's moving the other way]}+ \int v \qquad dt\]

Answer 46

That's what I'm talking about! That's great!

Answer 47

@IrishBoy123 35.667

Answer 48

No need for you to check, I use my calculator for that one :p

Answer 49

Fancy!

Answer 50

http://www.wolframalpha.com/input/?i=%5Cint_%7B0%7D%5E%7B3%7D+x%5E2-9x%2B18+dx+-%5Cint_%7B3%7D%5E%7B6%7D+x%5E2-9x%2B18+dx%2B%5Cint_%7B6%7D%5E%7B8%7D+x%5E2-9x%2B18+dx

Answer 51

are we all agreed?!?!

🎃

Answer 52

Yups only took 4hrs lel. Thanks for the help, really appreciate it. @IrishBoy123 @MaydayPaRAYde

Answer 53

:) That's awesome I'm Glad he was here to put you on the right track!

Answer 54

thanks you mayday and ephemera

🍓🍓🍓

Answer 55

You're Welcome :3