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Origin of terms in equations

  1. Jan 16, 2010 #1
    I'm curious as to the origin of the "1/2" in some of the basic equations in physics. For example, d=1/2at[tex]^{2}[/tex]. If d=vt, and v=at, then d=at[tex]^{2}[/tex], yet in reality we need the "1/2". Why?
  2. jcsd
  3. Jan 16, 2010 #2


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    Staff: Mentor

    In d = vt, "v" is the average velocity during the time interval between time 0 and time t.

    In v = at, "v" is the instantaneous velocity at time t.

    With constant acceleration starting from rest, the average velocity during the time interval between time 0 and time t is v/2 = at/2 (where "v" is again the final velocity at time t).
  4. Jan 16, 2010 #3
    In [tex]d = vt + d_0[/tex] your velocity is changing with time. The velocity in this equation should be the average velocity, call it v'.

    [tex]v' = \frac{v_0+v}{2} => d = (v')t[/tex]

    In [tex]v = at + v_0[/tex] you're assuming constant acceleration so the average, a' is equal to a.

    [tex]d = v't + d_0 = \frac{(v_0+v)t}{2} + d_0= \frac{(v_0 + at + v_0)t}{2} + d_0[/tex]

    So simplifying we get:

    [tex]d = d_0 + v_0 t + \frac{1}{2}at^2[/tex]
  5. Jan 16, 2010 #4
    Thanks alot, that makes sense now, being the average. So how does it arise in E=1/2mv[tex]^{2}[/tex]?
  6. Jan 16, 2010 #5
    Kinetic energy is for instantaneous velocity. Just thinking about it you should get a contradiction if you take an average energy vs. an average velocity due to the velocity being squared.
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