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Initial horizontal velocity of the boat

  1. Feb 19, 2006 #1
    in a water tank involving the launching of a small model boat,the model's initial horizontal velocity is 20ft/s and its horizontal acce varies linearly from -40ft/s^2 at t=0 to -6ft/s^2 at t=t(1) then remains equal to -6ft/s^2 until t=1.4s.Knowing that v=6ft/s when t=t(1), determine the value of
    t(1)

    pls help.
    the answer is 0.609s
     
  2. jcsd
  3. Feb 20, 2006 #2
    pls help.........
     
  4. Feb 20, 2006 #3
    Find the accn. as a function of time (Assume t(1) = T or something else).
    Then put it equal to dv/dt. Integrate. Put limits of time as (0,T) and velocity as (20,6).
     
  5. Feb 20, 2006 #4
    no,still can't get.somebody pls help
     
  6. Feb 20, 2006 #5

    VietDao29

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    Until here, the problem tells you that the initial horizontal velocity of the boat is 20 ft / s.
    What does the phrase varies linearly tell you? If you plot the graph of acceleration - time, then from t = 0 to t = t(1), do you get a straight line or what?
    I dunno, but I have a feeling that you don't need this.
    If v = 6ft / s. Can you find t(1)?
    Hint, you need to integrate [tex]\Delta v = \int \limits_{0} ^ {t(1)} a(t) dt[/tex].
    Hint: it's a right trapezoid. Can you find the area of it?
    Then you may also need to use the equation: [tex]v = v_0 + \Delta v[/tex]
    Can you go from here? :)
     
    Last edited: Feb 20, 2006
  7. Feb 20, 2006 #6
    what u mean is it -40t(1) = final velocity - initial velocity??
    40t(1) = 6ft/s - 20ft/s
     
  8. Feb 20, 2006 #7

    VietDao29

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    You are close, but it's neither 40t(1) nor -40t(1).
    The area under the graph of a(t) from 0 to t(1) is the area of a right trapezoid. Do you see why?
    I'll give you another big hint, how can one find the area of a right trapezoid? What are the length of the parallel sides, what's the height of that right trapezoid?
    Hopefully, you can go from here, right? :)
     
    Last edited: Feb 20, 2006
  9. Feb 20, 2006 #8
    is it 6(1.4-t(1) )??no,still can't find it.pls tell me
    thanx
     
  10. Feb 20, 2006 #9

    VietDao29

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    Okay, take out a pen and some paper, varies linearly means that if you plot an acceleration - time graph, you will get a straight line (that's what linear means).
    So at t = 0, yor acceleration is -40 ft / s (that means the line goes through the point (0; -40)), and at t = t(1), your acceleration is -6 ft / s (that means the line goes through the point (t(1), -6)). t(1) is some positive value, just choose it randomly.
    Now just connect the 2 point (0; -40), and (t(1), -6), you'll have a line. That line shows you the acceleration on the interval [0; t(1)], right? And it's a right trapezoid.
    The area of a trapezoid is:
    [tex]A = \frac{1}{2} (a + b) h[/tex]
    The parallel sides' lengths are 40, and 6. The height is t(1). Do you see why? Just look at your graph.
    But since the trapezoid is under the x-axis, so:
    [tex]\Delta v = \int \limits_{0} ^ {t(1)} a(t) dt = -A_{\mbox{trapezoid}}[/tex]. Do you know why?
    So you'll have:
    [tex]\Delta v = \int \limits_{0} ^ {t(1)} a(t) dt = -A_{\mbox{trapezoid}} = -\frac{1}{2} (40 + 6) t(1)[/tex].
    Now it's almost there (I almost show you the answer :smile:). Can you go from here? :)
    If you have any problem understanding anything from my post, just shout it out. :)
     
    Last edited: Feb 20, 2006
  11. Feb 20, 2006 #10
    oh.got it.thank you very much
     
  12. Feb 20, 2006 #11
    while for the second part,what is the position of the model at t=1.4s??
    what formula can i use
     
  13. Feb 21, 2006 #12

    VietDao29

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    Now, just do it step by step. You may want to express the velocity of the boat in terms of t, so you are looking for a function v(t). Then just integrate that, and you'll have the displacement of the boat.
    [tex]d = \int \limits_{0} ^ {1.4} v(t) dt[/tex]
    Now the problem is how can you find v(t)?
    Hint, you need to find a(t) first, then integrate that, and solve for v(t).
    Can you go from here?
     
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