anonymous
  • anonymous
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.
Calculus1
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anonymous
  • anonymous
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.
Calculus1
schrodinger
  • schrodinger
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anonymous
  • anonymous
Hint: \[v(t) = \frac{ dx(t) }{ dt } \] You'll have to integrate this over the given interval
anonymous
  • anonymous
I got all the way through it but not the right answer. I got -3/4 for my displacement
anonymous
  • anonymous
I did integrate it. I'm just not coming up with the correct answer for some reason

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anonymous
  • anonymous
You may have made an arithmetic mistake, because -3/4 doesn't sound right
anonymous
  • anonymous
where did you get the 9t? I got
anonymous
  • anonymous
When you integrate a constant, you get at+C so \[\int\limits 9 dt = 9t+C\]
anonymous
  • anonymous
you don't need a constant it's on a definite integral
anonymous
  • anonymous
I know, I'm just explaining the 9t
anonymous
  • anonymous
\[\int\limits_{0}^{3} 9 dt \implies 9t|_0^3\]
anonymous
  • anonymous
yes! okay awesome I fixed part a with 21/2
anonymous
  • anonymous
|dw:1443314364425:dw| some handy rules, just in case/ review
anonymous
  • anonymous
AWESOME that's so helpful thank you
anonymous
  • anonymous
Yw, so what did you get for your displacement now?
anonymous
  • anonymous
21/2
anonymous
  • anonymous
hmmm it told me it was correct
anonymous
  • anonymous
but I can't get part b
anonymous
  • anonymous
Oh sorry I kept using 9 when there is a 4 haha :)
anonymous
  • anonymous
hahaha okay :) now I got 87/10 for part b & they counted it wrong
anonymous
  • anonymous
Let me just fix that to show it again \[\int\limits\limits_{0}^{3} (5t-4) dt \implies (\frac{ 5t^2 }{ 2 }-4t)|_0^3\] ah there we go
anonymous
  • anonymous
The distance is a bit tricky, because we have to find the total interval so we have \[5t-4 = 0 \implies t = \frac{ 4 }{ 5 }\] so our integral should be \[\int\limits_{0}^{4/5} (5t-4)dt + \int\limits_{4/5}^{3} (5t-4)dt\] and note that distance is a scalar quantity so you will have to use absolute values for your integrals
anonymous
  • anonymous
At t =4/5 is really just when it changes the direction
anonymous
  • anonymous
OH oh my gosh that makes so much more sense
anonymous
  • anonymous
i got 73/10 this time.. still incorrect any suggestions?
anonymous
  • anonymous
\[|\int\limits_{0}^{4/5} (5t-4)dt| \implies |(\frac{ 5t^2 }{ 2 }-4t)|_0^{4/5} |\] the integral is the same your interval is just changed so here is the first one
anonymous
  • anonymous
The |...| represent absolute values
anonymous
  • anonymous
Yes! I did that
anonymous
  • anonymous
Ok what did you get for this integral?
anonymous
  • anonymous
\[(4t-\frac{ 5t ^{2} }{ 2 }) Integral (0, 4/5) + (\frac{ 5t ^{2} }{ 2 } - 4t) \int\limits (3, 4/5)\]
anonymous
  • anonymous
Why did you write \[4t-5t^2/2\]
anonymous
  • anonymous
because that's on the integral (0, 4/5)? maybe not?
anonymous
  • anonymous
Noo, why are you changing the integrand? It's the same we are just seeing where the velocity is negative and positive
anonymous
  • anonymous
okay so do I set the integral negative? hahaha I'm so sorry
anonymous
  • anonymous
No, we are finding the distance, which is the magnitude it cannot be negative
anonymous
  • anonymous
so we're just finding the velocity which means just plugging in the integral? & keeping the integral the same?
anonymous
  • anonymous
Nooo, 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
  • anonymous
v(t) = 5t-4 is a straight line, the line does not change, but the path does
anonymous
  • anonymous
If you meant put the first integral as \[- \int\limits_{0}^{4/5} (5t-4)dt\] that would work
anonymous
  • anonymous
OH okay let me try!
anonymous
  • anonymous
would my second integral be \[\int\limits_{0}^{3} (5t-4)\]
anonymous
  • anonymous
Noooo haha, I have already set up the integrals for you, hmm ok, I think you're not understanding the concept
anonymous
  • anonymous
I feel like I'm losing it more and more hahaha okay let me look this over again
anonymous
  • anonymous
\[\left| \int\limits_{0}^{4/5}(5t-4)dt \right|+\left| \int\limits_{0}^{4/5} (5t-4)dt\right|\] that's all there is!
anonymous
  • anonymous
HAHAHAHAHA OH MY GOSH
anonymous
  • anonymous
any ideas why it's not taking my answer of 3.2? sometimes it's picky
anonymous
  • anonymous
That's not right
anonymous
  • anonymous
Lets do this one at a time \[\int\limits_{0}^{4/5} (5t-4)dt\] what does this integral =? Remember to include absolute value
anonymous
  • anonymous
I got -1.6 but of course the absolute value is 1.6
anonymous
  • anonymous
the integral is \[\frac{ 5t ^{2} }{ 2 }-4t \int\limits (0,4/5)\]
anonymous
  • anonymous
Why is your notation like that, \[\int\limits\limits_{0}^{4/5} (5t-4)dt = (\frac{ 5t^2 }{ 2 }-4t)|_0^{4/5} = \left( \frac{ 5\left( \frac{ 4 }{ 5 } \right) ^2}{ 2 }-4(\frac{ 4 }{ 5 }) \right)-0 = |\frac{ 5 \times 16}{ 2 \times 25}-\frac{ 16 }{ 5 }| = \frac{ 8 }{ 5 }\] so that's good, that is 1.6
anonymous
  • anonymous
Now lets do the other integral \[\int\limits_{4/5}^{3} (5t-4)dt = \left( \frac{ 5t^2 }{ 2 }-4t \right)|_{4/5}^{3}\]
anonymous
  • anonymous
it's normally not like that I just struggle with plugging it into a computer, my apologies
anonymous
  • anonymous
\[\frac{ 121 }{ 10 }\] this integral gives the above, you can check
anonymous
  • anonymous
Now we add the two integrals \[\frac{ 8 }{ 5 } + \frac{ 121 }{ 10 } = \frac{ 137 }{ 10 }\]
anonymous
  • anonymous
when I plug 3 in, i get \[\frac{ 45 }{ 2 } - 12=21/2\] ...what is going on
anonymous
  • anonymous
\[\left( \frac{ 5(3)^2 }{ 2 } -4(3)\right) - \left( \frac{ 5\left( \frac{ 4 }{ 5 } \right)^2 }{ 2 } -4(\frac{ 4 }{ 5 })\right)\] this is the result of the second integral, you're forgetting about \[\frac{ 4 }{ 5 }\]
anonymous
  • anonymous
yes so this would be 21/2 - 8/5 = 89/10
anonymous
  • anonymous
No, it should be +8/5, the second term gives -8/5 remember? And you're already subtracting using FTC so a (-)(-) = +
anonymous
  • anonymous
We figured this out with the first integral
anonymous
  • anonymous
ohhhhhhhhhhhhhh

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