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V^2 is used in certain formulae. Physically, what does it mean?

  1. Jul 28, 2013 #1
    We are all familiar with E=MC2, although I don't know how that formula relates to the big bang. In any case, I suspect I don't have the math to understand it and it would probably take several pages to define it.

    Everyone is familiar with F=MA. What most people don't know is that if you multiply both sides of that equation by D (distance), you get FD = MV2 which is the mathematical equivalent of E= MC2.

    What I am curious about is what does that equation mean? I know, physically, what velocity (D/T) is, I know physically what acceleration (D/T2) is, but I don't know physically what V2is? Can anyone explain it in simple terms?
     
    Last edited: Jul 28, 2013
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  3. Jul 28, 2013 #2

    russ_watters

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    Staff: Mentor

    Welcome to PF!

    You have it there: E=FD (or W=FD). So V2 comes into pay in (kinetic) energy type equations. By itself V2 is nothing though; it is just a fragment of an equation.
     
  4. Jul 28, 2013 #3

    George Jones

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    With an appropriate factor of 1/2, this just a special case of the work-energy theorem, which is taught in every first-year physics course, and which I taught to my students last week. If a constant force (directed parallel to motion) acts on an object that starts from rest, and that moves a distance D, then the force does work FD on the object, and this is equal to the kinetic energy Mv^2/2 that the body picks up.
     
  5. Jul 29, 2013 #4
    Energy, momentum, force, time and distance

    A force acting on an object will accelerate the object in the direction of the action of the force.

    The product of the force and the time it acts is equal to the product of the mass of the object and its velocity at that specified time and is the object's momentum.

    Plot the graph of the object's momentum versus its velocity.

    For velocities which are small compared with lightspeed this graph will be linear. The area underneath it is the area of a triangle, which is half base times height.

    This is 1/2 x momentum x velocity which equals 1/2/mv^2.

    So, the kinetic energy of a moving body can be thought of as being equal to its momentum integrated with respect to its velocity which is not only interesting but also largely ignored by physics teachers!
     
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