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Ball released from rest at the top of slope

  1. Oct 29, 2010 #1
    1. The problem statement, all variables and given/known data

    In an experiment, a ball was released from rest at the top of a slope and rolled a distance of 2.0 m down the slope in 8.0 s. Calculate


    (a) its average speed in this time
    (b) the speed after 8.0s from rest
    (c) the acceleration of the ball

    Basically, I got the answer for a...

    Which is simply Distance Tot. divided by Time Taken

    = 2.0 m/8.0 s
    = 1/4 m/s

    The answer for b is 0.50 m/s. I have come to a conclusion that the average speed is x0.5 of the speed at that particular instance.
    (2x Avg. speed in that 8s = Speed of object in 8s)

    However, I don't get why this came to be. Someone please explain to it to me. Thanks

    For c... Simple also (after getting b)

    Which is Velocity/Time

    0.50 m/s divided by 8s = 0.0625 m/s 2

    Basically, All I need is an explanation for Part b.
     
  2. jcsd
  3. Oct 29, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The speed uniformly increases from 0 to some final speed Vf. So the average speed is:
    Vave = (0 + Vf)/2 = Vf/2.

    (If that's not convincing enough, you can also derive this fact more formally using standard kinematics equations.)
     
  4. Oct 29, 2010 #3
    Thanks :)
     
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