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Homework Help: Find position and velocity at specific time

  1. Sep 10, 2010 #1

    JJBladester

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    1. The problem statement, all variables and given/known data

    The acceleration of point A is a = -5.4sin(kt) ft/s2.

    k = 3 rad/s.

    When t = 0, x = 0 and v = 1.8 ft/s.

    Determine the position and velocity of point A when t = 0.5s

    Answers: x = 0.598 ft v = 0.1273 ft/s


    2. Relevant equations

    [URL]http://latex.codecogs.com/gif.latex?x=distance[/URL]

    [URL]http://latex.codecogs.com/gif.latex?\dot{x}=velocity[/URL]

    [URL]http://latex.codecogs.com/gif.latex?\ddot{x}=acceleration[/URL]

    [PLAIN]http://latex.codecogs.com/gif.latex?\dot{x}(t_{2})=\dot{x}(t_{1})+\int_{t_{1}}^{t_{2}}\ddot{x}dt [Broken] [Broken]

    [PLAIN]http://latex.codecogs.com/gif.latex?x(t_{2})=x(t_{1})+\int_{t_{1}}^{t_{2}}\dot{x}dt [Broken] [Broken]

    3. The attempt at a solution

    The velocity of point A at t = 0.5s is:

    [PLAIN]http://latex.codecogs.com/gif.latex?\dot{x}(t_{2})=\dot{x}(t_{1})+\int_{t_{1}}^{t_{2}}\ddot{x}dt [Broken] [Broken]

    [URL]http://latex.codecogs.com/gif.latex?\dot{x}(0.5)=1.8+\int_{0}^{0.5}[-5.4sin(kt)]dt=0.1273ft/s[/URL]


    To find the position of point A at t = 0.5s, I want to use this formula
    [PLAIN]http://latex.codecogs.com/gif.latex?x(t_{2})=x(t_{1})+\int_{t_{1}}^{t_{2}}\dot{x}dt [Broken] [Broken]

    But, since I don't have a formula for [tex]\dot{x}[/tex], I'm not sure how to go about solving for the position of point A at time t = 0.5s.

    So far, I have:

    [URL]http://latex.codecogs.com/gif.latex?x(t_{2})=\int_{0}^{0.5}\dot{x}dt[/URL]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 10, 2010 #2

    kuruman

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    Can you find a functional form for the velocity v(t) first? Use

    [tex]v(t)=\int^{t}_{0}a(t)dt[/tex]

    Once you have that, you can evaluate at any time. For the position, don't forget that x(t) is proportional to a(t). What is the constant of proportionality?
     
  4. Sep 10, 2010 #3

    JJBladester

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    So, if

    gif.latex?\dot{x}(t)=\int_{0}^{t}\ddot{x}dt.gif

    and

    [URL]http://latex.codecogs.com/gif.latex?\ddot{x}=-5.4sin(kt)[/URL]

    then

    [URL]http://latex.codecogs.com/gif.latex?\dot{x}=\int_{0}^{t}[-5.4sin(kt)]dt[/URL]

    What is the constant of proportionality?

    I know that:

    v=d/t

    a=v/t=d/t2

    so

    d = a*t2

    The ratio of distance to acceleration is then

    d/a = t2

    Is that in the right direction for what you're asking? Now how does that "contstant of proportionality" help me?
     
    Last edited by a moderator: Apr 25, 2017
  5. Sep 10, 2010 #4

    kuruman

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    [/URL]
    Can you do this integral?
    You don't know that. This expression works only if the acceleration is zero - not the case here.

    Not true for reasons explained above.

    If "constant of proportionality" doesn't ring a bell, integrate v(t) once more to get x(t), then evaluate at the appropriate times. Be sure your expressions reduce to the correct values at t = 0.
     
    Last edited by a moderator: Apr 25, 2017
  6. Sep 11, 2010 #5

    JJBladester

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    Okay, I have the answers now, but why are we not adding a constant of integration when integrating up from acceleration to velocity to position? Like a v0 and x0 term?

    __________________________
    My solution
    __________________________

    a(t) = -5.4sin(kt)
    v(t) = int(a(t)) = (5.4/k)cos(kt)
    x(t) = int(v(t)) = (5.4/k2)sin(kt)

    At 0.5s, v(t) = (5.4/3)cos(3*0.5) = 0.1273
    At 0.5s, x(t) = (5.4/32)sin(3*0.5) = 0.598

    So I got the answers which kind of makes me happy, but why doesn't the v(t) equation have a v0 term (constant of integration)? Likewise, why doesn't the x(t) equation have something like v0t + x0

    By the way, thanks for your help... And I appreciate you not just giving me the answers flat-out... Can't learn that way!
     
    Last edited: Sep 11, 2010
  7. Sep 11, 2010 #6

    kuruman

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    You add a constant of integration when you do math not physics. In physics there are always limits of integration that take into account the needed constants. More formally, you start from dv = a(t)dt and integrate with limits of integration as shown below

    [tex]\int^{v}_{v_0}dv=\int^{t}_{0}a(t)dt[/tex]

    Note that this describes physical reality and says that "at t = 0 the velocity is v0 and at t = t, it is v" which basically establishes v(t) and takes into account v(0). Do it formally this way and see what you get.
     
  8. Sep 11, 2010 #7

    JJBladester

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    Your explanation was fantastic. Here is what I've got.

    dynamics%20hw%201.jpg
     
  9. Sep 11, 2010 #8

    kuruman

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    Great. Now you are ready to understand the "constant of proportionality." Note that

    a(t) = -5.4 sin(kt)
    x(t) = (5.4/k2) sin(kt)

    In other words, x(t) = -(1/ k2) a(t), which says that the position at any time t is proportional to the acceleration at any time t. That's what I had in mind when I posted my first response to your question.
     
  10. Sep 11, 2010 #9

    JJBladester

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    Ok, so in this example, the constant of proportionality states a relationship between position and acceleration. Knowing that x(t) is -1/k2a(t), I can just find the acceleration at any point in time and multiply it by -1/k2 to find the position at that time. Neat.
     
  11. Sep 11, 2010 #10

    kuruman

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    And once you know x(t), you take its time derivative to find v(t). It's a shortcut.
     
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