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Concerning motion

  1. Mar 6, 2005 #1
    Hi, I was given the following question:

    Q: A body starts moving with an acceleration [itex]a=const[/itex]. After a time [itex]t[/itex] it has covered the distance [tex]\Delta x[/itex]. What is its speed at time [itex]t[/itex]?

    Answer: [tex]\frac{2\Delta x}{t}[/tex]

    Can somebody say how to arrive at this answer (both mathematically and intuitionally)? I know that [itex]\Delta x/t[/tex] would be the speed w/o acceleration, so the "2" must come from acceleration, but where does it come from?

    Can the answer be obtained from the formula [itex]v=v_0+at^2[/itex]?

    - Kamataat
     
  2. jcsd
  3. Mar 6, 2005 #2
    [tex]x-x_0=\frac{1}{2}(v_0+v_f)\Delta{t},\ \ v_0=0 m/s[/tex]
     
  4. Mar 6, 2005 #3
    Thank you!

    - Kamataat
     
  5. Mar 6, 2005 #4

    HallsofIvy

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    Another way: for CONSTANT acceleration, the "average" speed is just the average of the starting speed and ending speed: [itex]\frac{v_i+v_f}{2}[/itex]. Since the starting speed is 0, if the ending speed is v, this is [itex]\frac{v}{2}[/itex] so the distance covered in time [itex]\Delta t[/itex] would be [itex]\Delta x =\frac{v}{2}\Delta t}[/itex] so [itex]v= 2\frac{\Delta x}{\Delta t}[/itex].

    CAUTION: This is only true for CONSTANT acceleration!
     
    Last edited: Mar 6, 2005
  6. Mar 7, 2005 #5
    Halls, that's what my equation says :) [itex]1/2(v_0 + v_f) =[/itex] average velocity :smile:
     
  7. Mar 7, 2005 #6
    yup, i get it now. tnx
     
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