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Find how far a particle travels in 9 seconds

  1. Apr 9, 2006 #1
    (from rest)

    I need to find how far a particle travels in 9 seconds
    The only given is acceleration but I can find the antiderivative for velocity and position.


    a = 1 + 3 sqrt(t) mph/sec
    v = t + 2t^(3/2) + vo mph
    s = 1/2t^2 + 4/5*t^(5/2) + vo*t + so miles

    9sec == .0025 hour

    [tex]
    \int_0 ^{0.0025} |v(t)|dt
    [/tex]
    When I take the integral of velocity to find distance traveled, I don't get the correct answer. (should be 344 feet) I would assume that I'm doing something wrong with the units but I can't figure it out.
     
    Last edited: Apr 9, 2006
  2. jcsd
  3. Apr 9, 2006 #2
    What are you plugging in for V not and the starting position X not. Otherwise your integration seems correct.
     
  4. Apr 9, 2006 #3
    I'm just putting in 0 for both vo and so
    because from the info that I'm given it says it's from rest
     
  5. Apr 9, 2006 #4
    Well... if you solve it that way your getting miles. I suggest converting it to feet.
     
  6. Apr 9, 2006 #5
    It turns out to be .0178 ft with the above integral

    I also tried the upper limit as 9, which returns 240 ft
     
  7. Apr 9, 2006 #6
    Plug in .0025 hours for your equation: .5T^2 + (4/5)T^(5/2)

    You will then get miles since you integrated miles/hour.

    Since you want your answer in feet you need to convert your answer by multiplying how many feet are in a mile.
     
  8. Apr 9, 2006 #7
    Yep, 3.375x10^-6 miles x 5280 ft/mi = .0178 ft
     
  9. Apr 9, 2006 #8
    I'm getting .06525 miles when I plug in numbers.

    Although I'm plugging in seconds and then dividng by 3600 (Making mph into mps).
     
  10. Apr 9, 2006 #9
    I'm getting .06525 miles when I plug in numbers.

    Although I'm plugging in seconds and then dividng by 3600 (Making mph into mps).

    I tried your way and it does seem to be giving the wrong answer. I believe this is because you integrate acceleration in terms of seconds and then velocity in terms of hours. I believe you have to integrate in terms of one unit and thus divide the velocity equation by 3600. Giving you the same equation over 3600 and you plug in seconds.
     
  11. Apr 9, 2006 #10
    got it thanks
     
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