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Average acceleration.

  1. Aug 24, 2011 #1
    An Olympic sprinter can accelerate from rest out of the blocks to a top speed of about 11.5 m/s. This is accomplished in the first 15 m of a race. What is the average acceleration of the first sprinter?
    Given- vi- 11.5 m/s
    Distance-15m
    Vf-?
    t-?
    Which formula do I use to find these missing terms?
     
  2. jcsd
  3. Aug 24, 2011 #2

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
  4. Aug 24, 2011 #3
    Last edited by a moderator: Apr 26, 2017
  5. Aug 24, 2011 #4

    tiny-tim

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    s is always distance :wink:
     
  6. Aug 24, 2011 #5
    Oh haha thank you. :)
    Now how do you find the different times?
     
  7. Aug 24, 2011 #6

    tiny-tim

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    what times?? :confused:

    (i don't see any times in the question :redface:)
     
  8. Aug 24, 2011 #7
    You have to find acceleration, the formula is vf-vi over tf-ti right?
     
  9. Aug 24, 2011 #8

    tiny-tim

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    that formula certainly works if you know t

    but you only know vi vf and s, and there is a formula using that information :wink:
     
  10. Aug 24, 2011 #9
    d = vt + (1/2)at^2
    is that the formula?
     
  11. Aug 24, 2011 #10

    tiny-tim

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    no of course not :confused:
     
  12. Aug 24, 2011 #11
    Im not sure which formula to use...
     
  13. Aug 24, 2011 #12

    lightgrav

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    what's their _average_ speed during that 15m ?
    so how long should it have taken them to go 15m, at that average speed?
     
  14. Aug 25, 2011 #13

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
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