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Average velocity vs Final velocity

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Shouldn't average velocity multiplied by 2 be equal to final velocity. We were doing a lab and we calculated an average time of .54 seconds to travel 1.92 meters. Multiplying the average velocity by 2 gives an extremely different answer than vf=gt. I know there is probably some experimental error, but I would think it would be 2-3m/s off. Any help?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 22, 2011 #2

    Doc Al

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    Staff: Mentor

    For uniformly accelerated motion starting from rest, the average velocity will be half of the final velocity. Is that the situation?
     
  4. Nov 22, 2011 #3
    First of all, the final velocity is not generally the average velocity times 2. Imagine that the your object were to go from zero to 1 m/s really quickly, and then traveled at 1 m/s for the rest of the trip (if you think this is unrealistic, think of a scrap of paper dropped in the air- it gets up to its final speed quickly and stays there through the rest of the fall). Then the average velocity and the final velocity would both be close to 1 m/s.

    For your case, I'm assuming you're using something much denser than a piece of paper, and since you're using the equation vf=gt you're probably dealing with something dropped into a short free fall. In that case, just using vf=gt should give you something pretty close to the actual final velocity.
     
  5. Nov 22, 2011 #4
    Oh, and I forgot to mention, the final velocity is twice the average velocity in the special case that the object starts from rest and accelerates at constant velocity. If these conditions are met, then the final velocity is vf=a*T, the total distance is (1/2)a*T[itex]\stackrel{2}{}[/itex], and the average velocity is va=(1/2)a*T[itex]\stackrel{2}{}[/itex]/T=(1/2)a*T; so vf=2va (T denotes total travel time).

    So once again, assuming that your lab is a drop from free fall, the acceleration is close to constant "g" for a short fall, and the final velocity should be relatively close to twice the average.
     
  6. Nov 22, 2011 #5
    You are correct, it was freefall using a volleyball. Sorry, I should have included that information. I am assuming the numbers were so off due to experimental error with the timing. Thanks so much for your help.
     
  7. Nov 22, 2011 #6
    The way it turns out 1.92m/0.54s = 3.55m/s, multiply that by 2 and you get 7.1m/s. If you set vf = gt then it is 9.8 x .54 which results in 5.29m/s. This difference just seemed so large to me, but the first equation assumes everything is done very accurately, while the second equation does not really care what distance was travelled in that time. It assumes the distance travelled was 1.42m given h=1/2gt2.
     
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