anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
anonymous
  • anonymous
For the first one I'd take the integral of the function from 0 to 8 right?
anonymous
  • anonymous
@ganeshie8 Can you give this a look?

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anonymous
  • anonymous
well the instantaneous velocity at t=5 can be found by pluggin in t=5 the velocity function is the one given...
anonymous
  • anonymous
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 = ...
anonymous
  • anonymous
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.
anonymous
  • anonymous
And thanks for actually helping, after 2 hours you'd think there'd be more responses.
anonymous
  • anonymous
Yeah I'm sorry I'm Trying to explain slope.
anonymous
  • anonymous
First one should be 224 According to @MaydayPaRAYde the second part should be equal to -2
anonymous
  • anonymous
First on needs to be redone, just doesn't seem right to me.
anonymous
  • anonymous
one*
anonymous
  • anonymous
@IrishBoy123 do you have an idea?
IrishBoy123
  • IrishBoy123
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......💥
IrishBoy123
  • IrishBoy123
https://www.desmos.com/calculator/hons6cwrqv i meant "this"
anonymous
  • anonymous
Whoa..... hella confused now
anonymous
  • anonymous
Basically graph and solve?
anonymous
  • anonymous
Found my mistake for the first one. It'll be 26.66*1/8
anonymous
  • anonymous
#1 = 3.33 #2 = v= -2 s=2
IrishBoy123
  • IrishBoy123
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.
anonymous
  • anonymous
That's what I was planning to do, graph it to show the intervals and when it increase and decreases.
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
Oh... That's Helpful.
anonymous
  • anonymous
@IrishBoy123 what does the graph of the integral represent? displacement?
IrishBoy123
  • IrishBoy123
yes!!! displacement!! very important distinction
anonymous
  • anonymous
Ok so I would use that for the last part right?
anonymous
  • anonymous
Only problem now is the 4th one.
anonymous
  • anonymous
Ohhhhh
anonymous
  • anonymous
Derivative of velocity is equal to acceleration.
IrishBoy123
  • IrishBoy123
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.
IrishBoy123
  • IrishBoy123
sic *integration* integrating
IrishBoy123
  • IrishBoy123
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
IrishBoy123
  • IrishBoy123
gtg for a while you seem in command
anonymous
  • anonymous
Ight, just one more thing. Distance is always positive right?
anonymous
  • anonymous
Lel Pretty sure it is.
IrishBoy123
  • IrishBoy123
yes, it is a scalar.
IrishBoy123
  • IrishBoy123
ie no direction, like the boy band.....😁
anonymous
  • anonymous
Oh god.....that was terrible....
anonymous
  • anonymous
Still having trouble or did you bump it on accident?
anonymous
  • anonymous
Bumped for fun. I got this down.
anonymous
  • anonymous
@IrishBoy123 Can't I just split the intervals then add? Like from 0 to 3, 3 to 6, 6 to 8?
IrishBoy123
  • IrishBoy123
yup
anonymous
  • anonymous
That would give me distance correct?
IrishBoy123
  • IrishBoy123
yes
IrishBoy123
  • IrishBoy123
\[\text{Distance =} \int v + \int (-v) \text{[for when it's moving the other way]}+ \int v \qquad dt\]
anonymous
  • anonymous
That's what I'm talking about! That's great!
anonymous
  • anonymous
@IrishBoy123 35.667
anonymous
  • anonymous
No need for you to check, I use my calculator for that one :p
anonymous
  • anonymous
Fancy!
IrishBoy123
  • IrishBoy123
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
IrishBoy123
  • IrishBoy123
are we all agreed?!?! 🎃
anonymous
  • anonymous
Yups only took 4hrs lel. Thanks for the help, really appreciate it. @IrishBoy123 @MaydayPaRAYde
anonymous
  • anonymous
:) That's awesome I'm Glad he was here to put you on the right track!
IrishBoy123
  • IrishBoy123
thanks you mayday and ephemera 🍓🍓🍓
anonymous
  • anonymous
You're Welcome :3

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