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Why is distance = 1/2a(t)^2

  1. Jun 24, 2010 #1
    When I learn physics I like to visualize what is happening in my head.
    I can understand why this problem is multiplied by t^2; from my rational
    it is to get (m/s^2) just to meters. But why is this equation multiplied by
    one half? I do not understand why this happens. Can someone please explain
    the reasoning behind this?
  2. jcsd
  3. Jun 24, 2010 #2
    There are a couple of reasons that could be given. One involves calculus. More simply though, you can just use algebra. You know that for constant acceleration in one dimension:

    [tex]v = v_0 + at[/tex]

    [tex]v_{av} = \dfrac{v_0 + v}{2}[/tex]

    Now, if you have a body that accelerates but has a known average velocity, then its displacement is:

    [tex]x = x_0 + v_{av}t[/tex]

    So, you can plug in the average velocity formula:

    [tex]x = x_0 + \dfrac{v_0 + v}{2}t[/tex]

    But since we're considering constant acceleration, we can substitute the equation for v and get:

    [tex]x = x_0 + \dfrac{v_0 + (v_0 + at)}{2}t[/tex]

    [tex]x = x_0 + \dfrac{2v_0 + at}{2}t[/tex]

    And finally, after distributing, we end up with the familiar position function formula:

    [tex]x = x_0 + v_0t + \dfrac{1}{2}at^2[/tex]

    If you just set the initial position and velocity equal to zero, this reduces to the equation you cited. So you can see that the factor of 1/2 comes because we can conside the average velocity of a moving body by simply taking half the sum of its initial and present velocities.
  4. Jun 24, 2010 #3


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    Many first Physics courses do not show the math behind the formulas, you just have to take them as presented and learn to use them. When you get some more math under your belt, specifically calculus the equations you have memorized will be derived.

    The answer to your question is calculus. Since clearly you are not there yet we cannot explain very well. At present your observation that it makes the units work is an excellent one. Keep seeing things like that and you do well.
  5. Jun 24, 2010 #4
    Though there are many places in physics where calculus is indispensable, (e.g., deriving the electric potential from Coloumb's law), this doesn't happen to be one of them. In fact, arunma answered the OP nicely:

    This is worth understanding because it will come up again at least twice. First in the definition of kinetic energy, [itex]\frac12 mv^2[/itex], and second in the potential energy of a spring, [itex]\frac12 kx^2[/itex].
  6. Jun 24, 2010 #5
    OK, let's show you :)

    When you imagine v(t) graph, distans is representing as an area under the curve.

    In motion with constans velocity it is clear: v(t) is horizontal line and s=v*t (see picture below)

    In uniformly accelerated linear motion v(t)=a*t, so each point have coordinates (t, a*t).

    Then area under the curve it triangle. Area of the triangle equals 1/2*a*h where a is the length of the base of the triangle, and h is the height or altitude of the triangle.

    In this case s=1/2*t*(a*t) = 1/2*a*t^2 (see image)


    Attached Files:

    Last edited: Jun 24, 2010
  7. Jun 24, 2010 #6
    If you're accelerating at 10m/s^2 for 1 second: in that one second, you started out going 0m/s and you ended up going 10m/s. Your average speed would be right in between, at 5m/s.

    If your average speed is 5m/s for 1 second....your distance is 5m. And there's your half.
  8. Jun 24, 2010 #7
    This is helpful, but be careful: this argument only works when the initial speed is zero.
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