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

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Physical Significance of Numbers
  1. Significant numbers (Replies: 4)

  2. Significant Numbers (Replies: 5)