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Instantaneous Speed in a time interval

  1. Sep 22, 2012 #1
    The problem statement, all variables and given/known data
    In reference to Galileo's Inclined Plane experiment.

    2) What does it mean to say “The ball picked up the same amount of speed in each successive time interval.” Use symbols to write this sentence mathematically.

    5) For the ball falling vertically, if we believe that its speed changes at a constant rate (constant velocity), what is true about the instantaneous speed at the beginning, middle, and end of any given time interval.

    The attempt at a solution

    2) In order to pick up the same amount of speed in each identically-successive time interval, the acceleration must be uniform/constant. So in each time interval, the ball will cover the same amount of distance. D=ST or D/T=S.

    5) If the speed is increasing at a constant rate then in any given time interval: at the beginning, speed will be slowest; at the middle, speed will be greater; and at the end speed will be the greatest.


    I can't seem to think these through properly and I know that Velocity is merely speed with a direction and that it can be negative while speed must be positive. A zero acceleration would mean that falling ball is not speeding up at all a negative acceleration would mean that it is slowing down. But my answers seem wrong (or at the very least incomplete).
     
  2. jcsd
  3. Sep 22, 2012 #2
    This is not correct. But the question was not about the distance. It was about the speed. Focus on that.
     
  4. Sep 22, 2012 #3
    I suppose I did answer more in terms of distance.

    So if the ball picks up the same amount of speed in each successive time interval, then that means it would simply have a constant velocity in some direction, right? Which, therefore, means that it neither speeds up nor slows down, meaning that acceleration is zero.

    Am I more in the right "ballpark" with my answer now?
     
  5. Sep 22, 2012 #4

    Simon Bridge

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    If the ball "picks up speed" - how can it have constant velocity?
     
  6. Sep 22, 2012 #5
    Well that's one thing I managed to confuse myself with. I wasn't sure whether to interpret "picking up the same amount of speed" as increasing each distance using the same speed or equating the the distance traveled by compounding the speed from each time interval.

    But if it is adding the speed per interval than it would be a constant acceleration and I thought Voko said that this was incorrect?
     
  7. Sep 22, 2012 #6
    The first part of your answer was correct. If the ball gains the same amount of speed in each time interval, the acceleration is constant. But if the speed is increasing, does it follow that the distance travelled in each time interval would be the same?
     
  8. Sep 22, 2012 #7
    The premise was correct. Picking up the same amount of speed within each time interval does mean acceleration is constant. But your conclusion about distance does not follow. Nor is it required according to the problem statement. You need to express, mathematically, the speed, not the distance.
     
  9. Sep 22, 2012 #8

    Simon Bridge

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    If you were picking up the same amount of bananas in each time interval, you'd be increasing the amount of bananas you have by the same amount of bananas each time.

    If ni is the number of bananas after the ith time interval, then ni-ni-1=Δni is the number of bananas you picked up. For this to be the same each time is to say that Δni = Δn (a constant). If Δt=ti-ti-1 then we can say that Δn/Δt (the rate of change of bananas) is a constant.

    If you were picking up the same speed in each time interval then you'd be increasing the speed by the same amount of speed each time.
     
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