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anonymous
 one year ago
The velocity function, in feet per second, is given for a particle moving along a straight line.
v(t) = 5t − 4, 0 ≤ t ≤ 3
(a) Find the displacement.
(b) Find the total distance that the particle travels over the given interval.
anonymous
 one year ago
The velocity function, in feet per second, is given for a particle moving along a straight line. v(t) = 5t − 4, 0 ≤ t ≤ 3 (a) Find the displacement. (b) Find the total distance that the particle travels over the given interval.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hint: \[v(t) = \frac{ dx(t) }{ dt } \] You'll have to integrate this over the given interval

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got all the way through it but not the right answer. I got 3/4 for my displacement

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I did integrate it. I'm just not coming up with the correct answer for some reason

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You may have made an arithmetic mistake, because 3/4 doesn't sound right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where did you get the 9t? I got

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When you integrate a constant, you get at+C so \[\int\limits 9 dt = 9t+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you don't need a constant it's on a definite integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know, I'm just explaining the 9t

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{3} 9 dt \implies 9t_0^3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes! okay awesome I fixed part a with 21/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443314364425:dw some handy rules, just in case/ review

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0AWESOME that's so helpful thank you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yw, so what did you get for your displacement now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm it told me it was correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but I can't get part b

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh sorry I kept using 9 when there is a 4 haha :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hahaha okay :) now I got 87/10 for part b & they counted it wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let me just fix that to show it again \[\int\limits\limits_{0}^{3} (5t4) dt \implies (\frac{ 5t^2 }{ 2 }4t)_0^3\] ah there we go

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The distance is a bit tricky, because we have to find the total interval so we have \[5t4 = 0 \implies t = \frac{ 4 }{ 5 }\] so our integral should be \[\int\limits_{0}^{4/5} (5t4)dt + \int\limits_{4/5}^{3} (5t4)dt\] and note that distance is a scalar quantity so you will have to use absolute values for your integrals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0At t =4/5 is really just when it changes the direction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OH oh my gosh that makes so much more sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got 73/10 this time.. still incorrect any suggestions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{4/5} (5t4)dt \implies (\frac{ 5t^2 }{ 2 }4t)_0^{4/5} \] the integral is the same your interval is just changed so here is the first one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The ... represent absolute values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok what did you get for this integral?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(4t\frac{ 5t ^{2} }{ 2 }) Integral (0, 4/5) + (\frac{ 5t ^{2} }{ 2 }  4t) \int\limits (3, 4/5)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why did you write \[4t5t^2/2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because that's on the integral (0, 4/5)? maybe not?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Noo, why are you changing the integrand? It's the same we are just seeing where the velocity is negative and positive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so do I set the integral negative? hahaha I'm so sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, we are finding the distance, which is the magnitude it cannot be negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we're just finding the velocity which means just plugging in the integral? & keeping the integral the same?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nooo, we're finding the distance, and the reason we have split it into two integrals is because the integral 0 to 3 does not account for the negative and positive path of the particle, that is just the displacement, but for part b we want distance, that's total area travelled, so we have to find the change in velocity which is at \[t=4/5\] so we go from 0 to 4/5 and we add the positive area 4/5 to 3 for the total distance...I hope that makes sense now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0v(t) = 5t4 is a straight line, the line does not change, but the path does

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you meant put the first integral as \[ \int\limits_{0}^{4/5} (5t4)dt\] that would work

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would my second integral be \[\int\limits_{0}^{3} (5t4)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Noooo haha, I have already set up the integrals for you, hmm ok, I think you're not understanding the concept

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I feel like I'm losing it more and more hahaha okay let me look this over again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left \int\limits_{0}^{4/5}(5t4)dt \right+\left \int\limits_{0}^{4/5} (5t4)dt\right\] that's all there is!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0HAHAHAHAHA OH MY GOSH

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0any ideas why it's not taking my answer of 3.2? sometimes it's picky

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lets do this one at a time \[\int\limits_{0}^{4/5} (5t4)dt\] what does this integral =? Remember to include absolute value

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 1.6 but of course the absolute value is 1.6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the integral is \[\frac{ 5t ^{2} }{ 2 }4t \int\limits (0,4/5)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why is your notation like that, \[\int\limits\limits_{0}^{4/5} (5t4)dt = (\frac{ 5t^2 }{ 2 }4t)_0^{4/5} = \left( \frac{ 5\left( \frac{ 4 }{ 5 } \right) ^2}{ 2 }4(\frac{ 4 }{ 5 }) \right)0 = \frac{ 5 \times 16}{ 2 \times 25}\frac{ 16 }{ 5 } = \frac{ 8 }{ 5 }\] so that's good, that is 1.6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now lets do the other integral \[\int\limits_{4/5}^{3} (5t4)dt = \left( \frac{ 5t^2 }{ 2 }4t \right)_{4/5}^{3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's normally not like that I just struggle with plugging it into a computer, my apologies

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 121 }{ 10 }\] this integral gives the above, you can check

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now we add the two integrals \[\frac{ 8 }{ 5 } + \frac{ 121 }{ 10 } = \frac{ 137 }{ 10 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when I plug 3 in, i get \[\frac{ 45 }{ 2 }  12=21/2\] ...what is going on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left( \frac{ 5(3)^2 }{ 2 } 4(3)\right)  \left( \frac{ 5\left( \frac{ 4 }{ 5 } \right)^2 }{ 2 } 4(\frac{ 4 }{ 5 })\right)\] this is the result of the second integral, you're forgetting about \[\frac{ 4 }{ 5 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes so this would be 21/2  8/5 = 89/10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, it should be +8/5, the second term gives 8/5 remember? And you're already subtracting using FTC so a ()() = +

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We figured this out with the first integral
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