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V=0.5at^2 Why is it halved?

  1. Mar 9, 2005 #1

    eax

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    Lets take an example of someone travelling 10m/s^2 the person is travelling for 3seconds how far did he travel. Easy 10*3^2*0.5 = 45. Now with units it would look something like this

    10m/s^2 * 3s^2 * 0.5

    The 3s^2 makes the s^2 part irrevelant for 10m/s^2 thus becomes 90m. But why is it halved? All I know is that "it is the area under the the graph" so that doesn't proove enough. Why is it "the area under the graph"?

    Thanks in advance!
     
  2. jcsd
  3. Mar 9, 2005 #2

    dextercioby

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    Science Advisor
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    Because of the definition.The definition involves a definite integral which happens to be the area under the graph of the velocity.But velocity is constant which means its graph is a straight line.Therefore the area is very easy to compute,because it's the area of a right triangle...(and that involves the factor 1/2)

    Daniel.
     
  4. Mar 9, 2005 #3

    Doc Al

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    Staff: Mentor

    Another way to see where that 1/2 comes from is to think of it this way:
    Distance = (Ave Speed)*(Time)
    Ave Speed = (initial speed + final speed)/2 [since the acceleration is uniform]
    Ave Speed = (final speed)/2 [since it starts from rest]
    final speed = at [since acceleration is uniform]
    So... D = (at)/2 (t) = 1/2 a t^2.
     
  5. Mar 10, 2005 #4

    VietDao29

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    Hi,
    First, draw a graph of [itex]v = v_{0} + at[/itex] with the axis Ox is t and axis Oy is v.
    And think that, if within a [itex]\Delta t[/itex] time very very small, the speed does not change much. So if [itex]\Delta t \rightarrow 0[/itex], the speed will change less and less. So within that very small amount of time, the distance an object can go is: [itex]v_{t}\Delta t[/itex]. That distance can be represented by the area of a small rectangle with the height of [itex]v_{t}[/itex], and the width of [itex]\Delta t[/itex]. So in a lot of [itex]\Delta t[/itex] time, which sum up to be the amount of time the object goes, the distance the object will go is the area under the graph.
    Hope it help,
    Viet Dao,
     
  6. Mar 10, 2005 #5
    If you see a coefficient which is the reciprocal of a variable's power, you have good reason to suspect integration.
     
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