Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Speed with constant acceleration

  1. Apr 26, 2005 #1
    Using a Newton equation for position and substituting (v-v0)/a for t you can get the following formula:
    v^2= v0^2+2a(x-x0) where v^2 is supposedly speed. I understood speed to be tha magnitude of velocity which is the sqrt of the sum of the squares of the components (i.e. sqrt(vx^2+vy^2+vz^2)). I seem to be missing something.... how is speed then v^2? I am sure it is right in front of my face but I am just not seeing it.

    TIA
     
  2. jcsd
  3. Apr 26, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    v^2 is the speed squared; v is the speed.

    Realize that your formula ,v^2= v0^2+2a(x-x0), is for uniformly accelerated straight-line motion.
     
  4. Apr 26, 2005 #3
    Thank you. It just seems odd that if they would state that an equation is speed as a fucnction of position it would be much better to have that equation be v = sqrt(v0^2+2a(x-x0)). It can throw simple people like me off :).
     
  5. Apr 26, 2005 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Think of that equation as a relationship among all the different variables, not just as a formula for finding speed. Depending upon what information is given, you may wish to use it to solve for any of the variables. (And having a single v^2 is much nicer looking than that ugly square root! :smile: )
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Speed with constant acceleration
Loading...