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Linear motion equations

  1. Mar 10, 2012 #1
    how were these equations derived or found?

    x = xo + vot + 1/2at2


    I tried to use v=(xf-xi)/(tf-ti) and a=(vf-vi)/(tf-ti)

    but I don't know where the 1/2 comes from

    also this one:
    v2 = vo2 + 2a(xf-xi)
     
  2. jcsd
  3. Mar 11, 2012 #2

    Fredrik

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    The first one is what you get if you integrate the equation that describes constant acceleration twice. If the acceleration is constant (=a), we have
    $$x''(t)=a$$ This implies that there's a constant ##v_0## such that $$x'(t)=at+v_0.$$ The constant is denoted by ##v_0## because ##x'(t)## is the velocity at time t, and the result we just found implies that ##x'(0)=v_0##, so ##v_0## is the velocity at time 0. The result we found also implies that there's a constant ##x_0## such that
    $$x(t)=\frac 1 2 at^2+v_0t+x_0.$$ The constant is denoted by ##x_0## because it's equal to ##x(0)##, the position at time 0.

    The other one follows from the result for the velocity above, and
    $$v(t)-v_0=\int_0^t x''(t)dt=at.$$ These results imply that
    $$v(t)^2=(v_0+at)^2=v_0{}^2+a^2t^2+2v_0at=v_0{}^2+2a\bigg(\frac 1 2 at^2+v_0t\bigg)=(v_0)^2+2a(x(t)-x_0).$$

    I think the formula you wrote down only holds if ##v(t_i)=v_0##, so you may want to check your book to see if it says exactly what you wrote, or something slightly different.
     
    Last edited: Mar 11, 2012
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