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D=vt+1/2at^2 derivation

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  1. Apr 1, 2015 #1
    d=vt+1/2at^2
    Derive me this formula please... When I do I get 1/2(vt+at^2)...i know that I'm doing a mistake by not considering the initial velocity... So how do I correct it?
     
    Last edited by a moderator: Apr 1, 2015
  2. jcsd
  3. Apr 1, 2015 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF.

    Can you show us the steps of your derivation? That will help us find any errors.
     
  4. Apr 1, 2015 #3
    v=d/t ;d=vt
    a=d/t^2 ;d=at^2
    1/2d+1/2d=1/2vt+1/2at^2
    d=1/2(vt+at^2)
     
  5. Apr 1, 2015 #4

    berkeman

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    Thank you, that helps. Have you had any Calculus yet? Are you familiar with derivatives?
     
  6. Apr 1, 2015 #5
    Yup
     
  7. Apr 1, 2015 #6

    berkeman

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    Great. In that case, where did the bolded equation come from below?

     
  8. Apr 1, 2015 #7
    acceleration= velocity/time
     
  9. Apr 1, 2015 #8

    berkeman

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    Actually it's the change in velocity over time, or in calculus, a(t) = dv(t)/dt.

    But it's also possible to do the derivation for a constant acceleration using only triangles. Here's a YouTube video that helps to explain it:



    :smile:
     
  10. Apr 1, 2015 #9
    How can you say that displacement is area under the velocity curve?
     
  11. Apr 1, 2015 #10

    berkeman

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    That's from calculus:

    v(t) = dx(t)/dt

    a(t) = dv(t)/dt

    So when you integrate both sides of the first equation, you are effectively finding the "area under the curve"...

    [tex]x = \int{v(t) dt}[/tex]
     
  12. Apr 1, 2015 #11
    Ok thanks
     
  13. Apr 1, 2015 #12

    rcgldr

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    If acceleration is constant, then you can use algebra instead of calculus. Note that Δt means change in time, Δx means change in position, a = acceleration, v = velocity.

    initial velocity = v
    final velocity = v + a Δt
    average velocity = (initial velocity + final velocity) / 2 = ((v) + (v a Δt))/2 = v + 1/2 a Δt
    Δx = average velocity Δt = (v + 1/2 a Δt) Δt = v Δt + 1/2 a Δt^2
     
    Last edited: Apr 1, 2015
  14. Apr 2, 2015 #13
    it is not possible to write v = d/t but v=dx/dt and a = dv/dt
    then v(t) =∫a(t)dt =at + v , v is the initial velocity and the acceleration a is constant.
    it comes : x(t) =∫v(t)dt = 1/2 at^2 + vt + x(0), x(0) is the initial position assumed nill then x(0) =0

    finally the distance d is x(t) =d= vt + 1/2 a t^2
     
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