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Displacement formula question

  1. Oct 7, 2011 #1
    Hey, first post (:
    Was just wondering why the formula for distance is

    x = ½ at2

    if v = at

    and x = vt
    then why is it ½ when you substitute in a?

    x = distance
    v = velocity
    t = time
     
  2. jcsd
  3. Oct 7, 2011 #2

    berkeman

    User Avatar

    Staff: Mentor

    Are you familiar with integration and differentiation in calculus? That's the easist way to explain it.
     
  4. Oct 9, 2011 #3
    Nice post. Very good how you defined the variables

    x = vt
    means that there is no acceleration so a = 0, and you have to plug into the whole formula and not just part of it.
    ( ie with a=0 and using only half an equation, x = (1/2) at^2 becomes x = 0 )


    The actual forms of the equation are:
    x1-x0 = v0 x t with no acceleration ie a=0 Eq 0

    With constant acceleration
    x1-x0 = ( v0 + v1)/2 x t Eq 1
    where ( v0 + v1)/2 is the average velocity

    Also v1 = at + v0 Eq 2

    sustitute equation 1 into 2
    x1 - x0 = ( v0 + at + v0 )/2 x t
    or x1 - x0 = v0 t + (1/2) a t^2 Eq 3

    the final equation of for rectinear motion with constant acceleration is
    (v1)^2 - (v0)^2 = 2a(x1-x0) Eq 4

    Hope that helps.
    you have to plug into the whole formula - not just part of it.
     
  5. Oct 10, 2011 #4
    Thanks good sirs
     
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