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Kinematics Problem Final speed of the sprinter.

  1. Sep 16, 2007 #1
    Okay, first post... don't be too hard on me! Just got around to joining the forum even though i have been browsing for quite some time.


    1. The problem statement, all variables and given/known data

    A sprinter can accelerate with constant acceleration for 3.30 before reaching top speed. He can run the 100-meter dash in 10 s. What is his speed as he crosses the finish line?


    2. Relevant equations

    Kinematic Equations
    Sf = Si + Vi[tex]\Delta[/tex]t + .5a[tex]\Delta[/tex]t^2
    Vf^2 = Vi^2 + 2a[tex]\Delta[/tex]s
    Vf = Vi + a[tex]\Delta[/tex]t

    3. The attempt at a solution

    I have tried this problem using two methods that haven't gotten me very far. First I tried to look at the problem in two different steps: the first 3.30s and then the last 7.70s. I solved one equation for acceleration using the first interval:

    a =Vf1/3.30s

    But I realized I can't substitute this into equations referring to the second time interval.

    My second attempt I tried to look at the overall. I noted that the average velocity must be 10m/s over the whole 10s interval, but this attempt pretty much fizzled out at that point.



    This is an extra credit problem for my class, but I really want to go that extra mile (or 100m as the case may be). Please lend this noob a hand!
     
  2. jcsd
  3. Sep 16, 2007 #2

    cristo

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    Science Advisor

    Welcome to the forums!!

    You are correct in thinking that you need to split the motion into two parts. You want to find the final velocity, which will be equal to both the initial and final velocity of the second part (as there is no acceleration here). So, you can use v=d/t. You know the time, so now you need only work out distance. You know the total distance is 100m, so if you could work out the distance for the first part, and subtract it from 100m, then you will be able to solve the problem.

    So, which one of your equations can you use to find out the distance the sprinter covers whilst accelerating?
     
    Last edited: Sep 16, 2007
  4. Sep 16, 2007 #3
    So, I am getting Vf = (100-x)/7.70s

    For the first 3.30s I can use:

    Vf = Vi + a[tex]\Delta[/tex]t
    Vf= 0 + a(3.30)


    Hmmmm.... so then:

    3.30a = (100-x)7.70


    Is this the right track?
     
  5. Sep 16, 2007 #4
    Correction: 6.70s would be the second interval! D'oh!
     
  6. Sep 16, 2007 #5
    GOT IT!

    First I solved for x:

    X = Xi + Vi[tex]\Delta[/tex]t + .5a[tex]\Delta[/tex]t^2
    X = 0 + 0 + .5a(3.30)^2
    X = 5.445a

    Now substituting that into my earlier equation:

    3.30a = (100-x)6.70
    3.30a = (100-5.445a)6.70
    27.565a = 100
    a = 3.6278

    Hurray!

    Now it's a simple plug and chug:

    Vf = 3.30*3.6278
    Vf = 11.97

    ~12.0 m/s


    Thanks to cristo for pointing me in the right direction and making my first interactive experience a good one! See ya around the forums!
     
  7. Sep 17, 2007 #6

    cristo

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    Staff Emeritus
    Science Advisor

    You're welcome; well done!
     
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