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Physical Significance of Numbers

  1. Apr 1, 2003 #1
    Consider this case: A ball is allowed to fall freely with a constant acceleration of 10 m/s^2. What will its speed be after travelling a distance of 10 m?

    We get that:
    v[initial] = 0 m/s
    d = -5 m (downward direction is taken as negative)
    a = -10 m/s^2

    To find v[final], we will use the formula d = (v[final]^2-v[initial]^2) / (2a)

    Rearranging, we will get:

    v[final]^2 = 2ad + v[initial]^2

    By plugging the numbers, we will get:

    v[final]^2 = 100 m^2/s^2

    By mathematics, there are 2 solutions to this problem, that is v[initial][1] = +10 m/s and v[initial][2] = -10 m/s. I can only think of the situation where the ball has a downward velocity. Is the other answer (the one with +10 m/s velocity) possible/does it has a physical reality? If so, can anyone explain the situation? Thanks a lot!
  2. jcsd
  3. Apr 1, 2003 #2
    I think to obtain your first formula, the identity
    vf - vi = a*t
    is used.
    Next, you divide by a to get t, and plug that into d = a/2 * t^2.

    But strictly speaking, d, v, and a are vectors.
    So you are in fact not allowed to divide by a.

    You could work around this by saying you just talk about one component (the z-component) of each vector. But this implies that (vf-vi) has the same sign as a (i.e., minus).

    Of course, this is still true in the answer.
  4. Apr 1, 2003 #3
    u consider a stationary ball 'b' directly below the falling ball and calculate the velocity of 'b' in + with respect to free falling ball.

    Else, if free fall is not crucial, vertical take off of a rocket (containing the ball) with acceleration 10 m/s^2. v = u + at = final velocity of both rocket & ball.
    Last edited: Apr 1, 2003
  5. Apr 1, 2003 #4


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    The -10 m/s velocity occurs when you trace the evolution of the system going forward in time. Consider the trajectory of the ball before it had the 0 m/s velocity.

  6. Apr 2, 2003 #5
    Aaaah..... I got it! Thanks a lot Hurkyl!
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