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Velocity/Acceleration - A Question So Simple.

  1. May 19, 2004 #1
    Okay, I was reading "Understanding Physics" and I was understanding everything until I got to this:

    What is the velocity of a rolling ball at a paticular moment? Consider the first second of time. During that second the ball has been rolling at an average velocity of 2 ft/sec. It began that first second of time at a slower velocity. In fact, since it started at rest, the velocity at the beginning (after 0 seconds, in other words) was 0 ft/sec. To get the average up to 2ft/sec, the ball must reach correspondingly higher velocities in the second half of the time interval. If we assume that the velocity is rising smoothly with time, it follows that if the velocity at the beginning of the time interval was 2 ft/sec less than average, then at the end of the time interval (after one second), it should be 2 ft/sec more than average, or 4 ft/sec.

    What I don't understand is how did was 4 ft/sec determined.
  2. jcsd
  3. May 19, 2004 #2


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    The average velocity for the time interval was 2 m/s, at the begining of the time interval the velocity was 0 m/s, to average 2m/s it must be traveling faster then the average at the end of time interval, in fact it must be as much above average at the end as it was below average at the begining. It was 2m/s below average at the start of the time interval so it must be 2m/s above average at the end of the interval, thus 4m/s.
  4. May 19, 2004 #3


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    The stuff you quoted was very strange. Anyhow, I think they mean "velocity is rising linearly with time." If something goes at an average of 2 ft/sec for one second, then it travels 2 ft. Now, in this case, it's not travelling at a constant velocity, it starts at zero velocity, and has a constant acceleration such that at the end of one second, it has travelled an average of 2 ft/sec. So, what do we know?

    [tex]\vec{a}\ =\ ?[/tex]
    [tex]\vec{v}_{init}\ =\ 0\ ft/s\ [forward][/tex]
    [tex]\vec{\Delta d}\ =\ 2\ ft\ [forward][/tex]
    [tex]\Delta t\ =\ 1\ s[/tex]

    Solve for acceleration using the formula:

    [tex]\Delta d\ =\ v_{init}\Deltat\ +\ \frac{1}{2}a\Delta t^2[/tex]
    [tex]a\ =\ 4\ \frac{ft}{s^2}[/tex]

    Now, find the final velocity using the formula:

    [tex]v_{final}^2\ =\ v_{init}^2\ +\ 2a\Delta d[/tex]
    [tex]v_{final}\ =\ 4\ \frac{ft}{s}[/tex]

    If you want to be technical:

    [tex]\vec{v}_{final}\ =\ 4\ \frac{ft}{s}\ [forward][/tex]
  5. May 20, 2004 #4
    Integral & AKG thanks for thank-you for the explanations, they both were extremely helpful. I understand what the text was trying to explain now.
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