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Equation of Gravity's Acceleration Misunderstanding

  1. May 16, 2013 #1
    Hello physics forum!

    As a youngster, it is a great honor to be surrounded by innocent minds. For this, thank you.

    Now, my question:

    I understand that the equation for vertical displacement caused by gravity is "y=0.5gt^2" or "y=-4.9t^2" in meters or"y=-16t^2" in feet. I have yet to understand why where is a 0.5 in the equation. It is to my understanding that if an object moves at for simplicity, 10 m/s, its displacement will increase by 10 m per second in the same direction. For example, an object in space throw initially at 5 m/s will travel 5 m in the first second, then 10 m away from the initial point in 2 seconds, and so forth. Gravity is a constant acceleration, by -9.8 m/s^2 or -32 ft/s^2. The velocity changes by these rates, depending on which unit. An object dropped will travel, in one second, -9.8 m/s downward, then -19.6 m/s, then -29.4 m/s. But when dealing with displacement, 0.5 is introduced...why?
  2. jcsd
  3. May 16, 2013 #2


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    It has to do with figuring out displacement based on a changing velocity.

    For example (I am using feet for simplicity), if you hold an object and drop it, the initial velocity is zero, but after one second, the object has accelerated under the influence of gravity to 32 ft/s. How far has the object traveled in that same 1 second? Since the initial velocity is 0 and the velocity at t = 1 s is 32 ft/s, assuming a constant change in velocity in between, it would be reasonable to say that the distance traveled would be the average velocity multiplied by the duration, or:

    dist = Vavg * delta T

    In this case, delta T = 1 sec., and Vavg = (Vi + Vf) / 2

    Plugging in what we have calculated so far:

    dist = [(0 + 32) / 2] ft/s * 1 s = 16 feet.

    For t = 1 s to 2 s, a similar calculation is followed, but vi = 32 ft/s

    A similar result can be obtained using calculus.
  4. May 16, 2013 #3
    Oh! I understand!

    But if the velocity is constant as in zero acceleration is involved, such as (theoretically) in space, the displacement is the same as the velocity each second?
  5. May 16, 2013 #4


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    If your acceleration is zero, then the .5 goes away. The full equation for displacement is:

    x= .5 g t2 + V0t + X0
  6. May 16, 2013 #5
    In that equation, I presume that the output "x" is only for one dimension and can be used for both horizontal and vertical? And that x0 is the initial point that you begin from?
  7. May 16, 2013 #6


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    Just like v0 is the initial velocity, x0 is the initial displacement.

    The equations of motion can be applied in all three coordinate directions. You will notice that there are quite a few projectile motion problems in the Physics Homework section, and a lot of these problems involve motion in two directions.
  8. May 17, 2013 #7


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    Welcome to PF, Greatness.

    Usually -- but not always -- the initial displacement x0 is taken to be zero in intro physics problems.
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