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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.

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  1. anonymous
    • one year ago
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    Hint: \[v(t) = \frac{ dx(t) }{ dt } \] You'll have to integrate this over the given interval

  2. anonymous
    • one year ago
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    I got all the way through it but not the right answer. I got -3/4 for my displacement

  3. anonymous
    • one year ago
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    I did integrate it. I'm just not coming up with the correct answer for some reason

  4. anonymous
    • one year ago
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    You may have made an arithmetic mistake, because -3/4 doesn't sound right

  5. anonymous
    • one year ago
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    where did you get the 9t? I got

  6. anonymous
    • one year ago
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    When you integrate a constant, you get at+C so \[\int\limits 9 dt = 9t+C\]

  7. anonymous
    • one year ago
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    you don't need a constant it's on a definite integral

  8. anonymous
    • one year ago
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    I know, I'm just explaining the 9t

  9. anonymous
    • one year ago
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    \[\int\limits_{0}^{3} 9 dt \implies 9t|_0^3\]

  10. anonymous
    • one year ago
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    yes! okay awesome I fixed part a with 21/2

  11. anonymous
    • one year ago
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    |dw:1443314364425:dw| some handy rules, just in case/ review

  12. anonymous
    • one year ago
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    AWESOME that's so helpful thank you

  13. anonymous
    • one year ago
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    Yw, so what did you get for your displacement now?

  14. anonymous
    • one year ago
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    21/2

  15. anonymous
    • one year ago
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    hmmm it told me it was correct

  16. anonymous
    • one year ago
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    but I can't get part b

  17. anonymous
    • one year ago
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    Oh sorry I kept using 9 when there is a 4 haha :)

  18. anonymous
    • one year ago
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    hahaha okay :) now I got 87/10 for part b & they counted it wrong

  19. anonymous
    • one year ago
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    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

  20. anonymous
    • one year ago
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    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

  21. anonymous
    • one year ago
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    At t =4/5 is really just when it changes the direction

  22. anonymous
    • one year ago
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    OH oh my gosh that makes so much more sense

  23. anonymous
    • one year ago
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    i got 73/10 this time.. still incorrect any suggestions?

  24. anonymous
    • one year ago
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    \[|\int\limits_{0}^{4/5} (5t-4)dt| \implies |(\frac{ 5t^2 }{ 2 }-4t)|_0^{4/5} |\] the integral is the same your interval is just changed so here is the first one

  25. anonymous
    • one year ago
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    The |...| represent absolute values

  26. anonymous
    • one year ago
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    Yes! I did that

  27. anonymous
    • one year ago
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    Ok what did you get for this integral?

  28. anonymous
    • one year ago
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    \[(4t-\frac{ 5t ^{2} }{ 2 }) Integral (0, 4/5) + (\frac{ 5t ^{2} }{ 2 } - 4t) \int\limits (3, 4/5)\]

  29. anonymous
    • one year ago
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    Why did you write \[4t-5t^2/2\]

  30. anonymous
    • one year ago
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    because that's on the integral (0, 4/5)? maybe not?

  31. anonymous
    • one year ago
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    Noo, why are you changing the integrand? It's the same we are just seeing where the velocity is negative and positive

  32. anonymous
    • one year ago
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    okay so do I set the integral negative? hahaha I'm so sorry

  33. anonymous
    • one year ago
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    No, we are finding the distance, which is the magnitude it cannot be negative

  34. anonymous
    • one year ago
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    so we're just finding the velocity which means just plugging in the integral? & keeping the integral the same?

  35. anonymous
    • one year ago
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    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

  36. anonymous
    • one year ago
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    v(t) = 5t-4 is a straight line, the line does not change, but the path does

  37. anonymous
    • one year ago
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    If you meant put the first integral as \[- \int\limits_{0}^{4/5} (5t-4)dt\] that would work

  38. anonymous
    • one year ago
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    OH okay let me try!

  39. anonymous
    • one year ago
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    would my second integral be \[\int\limits_{0}^{3} (5t-4)\]

  40. anonymous
    • one year ago
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    Noooo haha, I have already set up the integrals for you, hmm ok, I think you're not understanding the concept

  41. anonymous
    • one year ago
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    I feel like I'm losing it more and more hahaha okay let me look this over again

  42. anonymous
    • one year ago
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    \[\left| \int\limits_{0}^{4/5}(5t-4)dt \right|+\left| \int\limits_{0}^{4/5} (5t-4)dt\right|\] that's all there is!

  43. anonymous
    • one year ago
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    HAHAHAHAHA OH MY GOSH

  44. anonymous
    • one year ago
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    any ideas why it's not taking my answer of 3.2? sometimes it's picky

  45. anonymous
    • one year ago
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    That's not right

  46. anonymous
    • one year ago
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    Lets do this one at a time \[\int\limits_{0}^{4/5} (5t-4)dt\] what does this integral =? Remember to include absolute value

  47. anonymous
    • one year ago
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    I got -1.6 but of course the absolute value is 1.6

  48. anonymous
    • one year ago
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    the integral is \[\frac{ 5t ^{2} }{ 2 }-4t \int\limits (0,4/5)\]

  49. anonymous
    • one year ago
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    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

  50. anonymous
    • one year ago
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    Now lets do the other integral \[\int\limits_{4/5}^{3} (5t-4)dt = \left( \frac{ 5t^2 }{ 2 }-4t \right)|_{4/5}^{3}\]

  51. anonymous
    • one year ago
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    it's normally not like that I just struggle with plugging it into a computer, my apologies

  52. anonymous
    • one year ago
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    \[\frac{ 121 }{ 10 }\] this integral gives the above, you can check

  53. anonymous
    • one year ago
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    Now we add the two integrals \[\frac{ 8 }{ 5 } + \frac{ 121 }{ 10 } = \frac{ 137 }{ 10 }\]

  54. anonymous
    • one year ago
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    when I plug 3 in, i get \[\frac{ 45 }{ 2 } - 12=21/2\] ...what is going on

  55. anonymous
    • one year ago
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    \[\left( \frac{ 5(3)^2 }{ 2 } -4(3)\right) - \left( \frac{ 5\left( \frac{ 4 }{ 5 } \right)^2 }{ 2 } -4(\frac{ 4 }{ 5 })\right)\] this is the result of the second integral, you're forgetting about \[\frac{ 4 }{ 5 }\]

  56. anonymous
    • one year ago
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    yes so this would be 21/2 - 8/5 = 89/10

  57. anonymous
    • one year ago
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    No, it should be +8/5, the second term gives -8/5 remember? And you're already subtracting using FTC so a (-)(-) = +

  58. anonymous
    • one year ago
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    We figured this out with the first integral

  59. anonymous
    • one year ago
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    ohhhhhhhhhhhhhh

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