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Converting Acceleration to Velocity

  1. Feb 25, 2007 #1
    1. The problem statement, all variables and given/known data
    Is velocity equivalent to the square root of accelleration?

    2. Relevant equations
    V = d/t

    3. The attempt at a solution

    A = accelleration, expressed in meters per second squared

    V = distance (m) divided by time (s)

    So V = sqrt A ?

    In the context of the question: How long does it take an object with a mass of 100g (.1kg) mass to travel 10 cm (.1m), with a force (assumed to be constant) of 0.0000000132944 N.
    Last edited: Feb 25, 2007
  2. jcsd
  3. Feb 25, 2007 #2

    Doc Al

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    Staff: Mentor

    Uh... no.

    If velocity were the square root of acceleration, it would have units of square root of distance divided by time.

    Time for you to review basic kinematics. Find an expression that relates distance with time for a given acceleration.
  4. Feb 25, 2007 #3
    Precisely, acceleration is the derivitive of velocity, expressed in m/s per second. Its the rate of change of velocity, or V/t. That means V = At

    Your question relates distance (not velocity) to acceleration, in which case you need to find the constant acceleration and then plug it into the kinematic equation D = vo + .5at^2
  5. Feb 25, 2007 #4
    If possible, could someone direct me to a website which provides an introduction (prefferably oriented towards beginners with very little prior knowledge of the subject or complementary subjects) to the kinematics involved in this equation?
  6. Feb 25, 2007 #5


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    Science Advisor

    Don't you have a text book that defines acceleration?
  7. Feb 25, 2007 #6

    Doc Al

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    Staff: Mentor

    Last edited by a moderator: May 2, 2017
  8. Feb 25, 2007 #7
    I wish I had a textbook for this... but alas, it is not covered in the Ontario Grade 8 curriculum, as such a physics textbook would be rather unneccessary, as it is not taught...
  9. Feb 25, 2007 #8
    Last edited by a moderator: May 2, 2017
  10. Feb 25, 2007 #9
    Having found the appropriate kinematic equation, I have calculated the following:

    d = Vi*t + 0.5*a*t^2

    = 0m/s*t + 0.5*a*t^2

    = a*t^2/2

    2d= a*t^2

    2(0.1m)/a = t^2

    0.2/0.000000132944m/s^2 = t^2

    sqrt 1 504 392.827055000617 = t

    t = 1 226.5369244 s

    Is the value for t, one thousand two hundred twenty - six seconds correct, and was the equation used correctly?
  11. Feb 26, 2007 #10

    Doc Al

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    Staff: Mentor

    That looks fine. A few comments:

    Express numbers such as the force using exponential notation (like 1.32944 × 10^-8 N); they are much easier to read that way instead of a string of zeroes.

    Take care with significant figures. It looks like your values for mass and distance only have one significant figure--yet your answer has around 20. Not very realistic! (Just because the calculator carries all those digits doesn't mean you should keep them in your answer.)
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