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How to calculate acceleration per second

  1. Aug 6, 2007 #1
    the law: a1 = G(m2/r*r) to calculate acceleration with gravity, gives the m/s*s.
    (i don't know how to type the square sign)
    but how can u calculate how many m/s an object gains in a certain time?
     
  2. jcsd
  3. Aug 6, 2007 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF. You use the equations of "kinematics", which are the basic equations of motion. For a constant acceleration (like gravity), the equations are fairly simple algebraic equations. If the acceleration is varying (like an accelerating rocket that gets lighter as the fuel burns off), then you use the calculus to calculate the motion of the object.

    Here's a background page with basic information from wikipedia.org:

    http://en.wikipedia.org/wiki/Kinematics
     
  4. Aug 7, 2007 #3
    ah thank you, i guess i was looking in the wrong direction.

    so to calculate it i guess i would have to use this formula?
    uf(small f) = ui(small i) + at

    is this true? because i dont really know how to calculate it. i need to have a list of its speed every 1/30th of a second
    so like:
    0kmph
    0.005kmph
    0.0020kmph
    etc
    i just don't know how to do that with this formula
     
    Last edited: Aug 7, 2007
  5. Aug 7, 2007 #4
    [tex]u_f=u_i+at[/tex] is an equation for constant acceleration (like gravity)
    where [tex]u_i[/tex] is your initial velocity; a is the rate of acceleration; t is time in seconds; and [tex]u_f[/tex] is your final[/tex] velocity.

    So, if you know the velocity an object is starting at; you know its rate of acceleration; you can simply plug in values for t (e.g. 1/30, 2/30...)
     
  6. Aug 7, 2007 #5
    but i dont know what "at" is.
    t is tiem and a is acceleration ok. but a is m/s*s and t is seconds, right?
    so that would mean: ms/s*s*s and i dont know what that is, so im probally doing it wrong, what is it?
     
  7. Aug 7, 2007 #6

    berkeman

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    Staff: Mentor

    When you multiply quantities, you multiply the units in the same way, and that gives you the units of the final quantity. In all equations, the units of the left hand side (LHS) must match the units of the right hand side (RHS), and the units of each quantity in an addition must match.

    So in the equation:

    [tex]v_f = v_i + a t[/tex]

    the units of v are [tex]\frac{m}{s}[/tex],

    the units of a are [tex]\frac{m}{s^2}[/tex],

    and the units of t are seconds [tex]s[/tex].

    When you multiply a * t, you get units of [tex]s \frac{m}{s^2} = \frac{m}{s}[/tex],

    which matches the units of [tex]v_f [/tex] and [tex]v_i [/tex]

    In fact, carrying units along with your calculations and checking for their consistency at each algebraic step is an important trick for helping to avoid making math errors in the middle of long calculations.

    Does that make sense now?
     
  8. Aug 7, 2007 #7
    yes, thank you. but my uncle told me that gravitational acceleration on earth is 9.8m/s(2). and that it means that the first second it will move 9.8m/s(in vacuum and in the second second ca 100m/s this is not true right?
     
  9. Aug 7, 2007 #8

    berkeman

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    Staff: Mentor

    That's a little too simplistic, but it's along the correct lines. You can't quantize the situation each second -- that introduces approximation errors. Instead, look at the equation for the position of a dropped object versus time:

    [tex]x = x_0 + v_0 t + \frac{a t^2}{2}[/tex]

    For a dropped object, call the initial position 0 and the initial velocity 0, so the first two terms on the RHS go away. Then you are left with the dropped distance x versus time like this:

    [tex]x = \frac{a t^2}{2}[/tex]

    So now you can calculate how far you go each second. After 1 second, you have dropped an x distance of

    [tex]9.8 \frac{m}{s^2} * {(1s)}^2 = 9.8 m[/tex]

    After 2 seconds, you have dropped a total x distance of

    [tex]9.8 \frac{m}{s^2} * {(2s)}^2 = 39.2 m[/tex]

    After 3 seconds,....... and so on. So once you have calculated how far the ball drops total for each time, you can go back and figure out how far it dropped in the first second (9.8m), and then how far in the second second (39.2-9.8m), etc.
     
    Last edited: Aug 7, 2007
  10. Aug 7, 2007 #9
    so i can't calculate it's speed after 1/30th of a second and then using that as a base calculate its speed after 2/30th of a second etc?
     
  11. Aug 7, 2007 #10

    berkeman

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    Staff: Mentor

    Yes, you can do it that way. Just use the full form of the velocity equation:

    [tex]v = v_0 + a t[/tex]

    If the object is dropped from rest, then Vo is 0. If you figure out what V is after 1/30 of a second using the equation, you then call that time zero, and use that velocity as Vo and use the equation again. Hopefully I'm not confusing you with all of this. I was a bit confused with the mixing of your questions about position and velocity. Are we square now?
     
  12. Aug 7, 2007 #11

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Every things fine as long as the acceleration is a constant. If the acceleration varies with time, then the velocity is the integral of the acceleration with respect to time. Since gravitional force, and so acceleration, varies with respect to the distance (thus with respect to time if the distance is varying), Newton and Leibniz were force to invent calculus to deal with problems like that!
     
  13. Aug 7, 2007 #12
    no i understand everything now I made that equation at first aswel, but after what you said i had doubts if it was correct. thank you very much.
     
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