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How to integrate velocity

  1. Nov 10, 2015 #1
    If i have an equation: h = ½at2
    And:
    40 = -4.9t2 → 0 = -4.9t2 - 40

    If: v = ∫a

    then how do i find the velocity given acceleration of -9.8 m/s2?
     
  2. jcsd
  3. Nov 10, 2015 #2

    berkeman

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    Staff: Mentor

    You are missing some constants in those equations. Are you familiar with the full equations of motion given a constant acceleration a?

    v(t) = ?
    x(t) = ?
     
  4. Nov 10, 2015 #3

    Well Im only a Sophomore and haven't had either Calculus nor Physics..
    However, v(t) is what i need to know.
    What is x(t)?
     
  5. Nov 10, 2015 #4

    berkeman

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    x(t) is the linear distance travelled. It looks like you are wanting z(t) or h(t) height, but it doesn't matter, you still use the same equations.

    See this Hyperphysics page for example:

    http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#motcon

    It lists the equations for v(t) and x(t) or h(t). Although, they have left off one term -- there should be an x(0) term on the righthand side of the equation for x(t).

    Does that help?
     
  6. Nov 10, 2015 #5
    Thanks,

    So how do i find the velocity at a given time? Do I need all of those variables?
    If height is 40:
    40=0.5(9.8)(t^2) ........ We can solve this for time, however, how do I find the velocity?
    Sorry, if I'm missing something thats obvious.

    So does v = at?
     
  7. Nov 10, 2015 #6

    berkeman

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    Your problem statement is still not clear to me. We need to know the initial conditions h(0), v(0) in order to do any calculations. To summarize, you will use equations like these (I'm going to use y(t) for the height);

    [tex]v(t) = v(0) + a*t[/tex]

    [tex]y(t) = y(0) + v(0)*t + \frac{1}{2}a*t^2[/tex]
     
  8. Nov 10, 2015 #7
    Okay.... Thank you!

    Sorry for the ignorant questions.
     
  9. Nov 10, 2015 #8

    berkeman

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    Your welcome. They are not ignorant questions -- you are learning. :smile:
     
  10. Nov 10, 2015 #9

    berkeman

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    BTW, it's important that you get the signs right for whatever coordinate system you define. So if y(t) is positive up, then the acceleration in the y direction is -9.8m/s^2 as you have already said. The initial position and velocity will have signs as well in the general problems.
     
  11. Nov 10, 2015 #10
    Okay! Thank you!
     
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