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Olympic class runner

  1. Jan 15, 2007 #1
    1. Careful measurements have been made of Olympic sprinters in the 100 meter dash. A simple but reasonably accurate model is that a sprinter accelerates at 3.8 m/s2 for 3.26 s, then runs at constant velocity to the finish line.
    (a) What is the race time for a sprinter who follows this model?

    (b) A sprinter could run a faster race by accelerating faster at the beginning, thus reaching top speed sooner. If a sprinter's top speed is the same as in part a, what acceleration would he need to run the 100 meter dash in 9.61 s?

    (c) By what percent did the sprinter need to increase his acceleration in order to decrease his time by 1%?




    2. Relevant equations
    Vf=Vi+at
    Sf=Si +Vi*t+.5*a*t^2




    3. The attempt at a solution

    ok i am able to get part (a):

    i started by finding the velocity for the rest fo teh race after the acceleration:
    Vf=0+((3.8)(3.26))=12.388 m/s

    then i found how far he traveled during the acceleration:
    Sf=0+0+.5((3.8)(3.26))=20.19m

    next i found how much time it took to complete the race from t=3.26s:
    100=20.19+(12.388)t+.5(0)t^2
    100-20.19=12.388t
    t=6.44s

    finally:
    t(total)= 3.26+6.44= 9.7s

    now i have been looking at the last two parts and i have no idea where to start.
     
  2. jcsd
  3. Jan 15, 2007 #2
    What have you tried so far for parts b and c ?
    To start you off on b, look at the given information. You have total distance, total time and maximum velocity. The only factors you do not know directly are the acceleration, the acceleration phase time and the constant velocity time.
    Thus you have 3 eqns and 3 unknowns, I think you know what to do now :)
     
  4. Jan 15, 2007 #3
    Looking at your final equation you have:

    100 = 20.19 + 12.388t + .5(0)t^2

    In this new situation the only thing we are changing is the 20.19 part because we don't know how far he goes in the first leg, so that part can become X lets say

    100 = x + 12.388t

    Now we just gotta think, what exactly is x. Well look how you calculated it previously

    Sf=0+0+.5((3.8)(3.26)^2)=20.19m

    Since we don't know the 3.8 or 3.26 we can just fill them as A and [itex]t_1[/itex] respectively ([itex]t_1[/itex] because it's not the same time as the time variable thats already in the equation).

    [tex] 100 = (.5)(a)(t_1)^2 + 12.388t [/tex]

    That is an equation with 3 unknowns. The rest is is just using known information to eliminate variables. See if you can solve it from there
     
    Last edited: Jan 15, 2007
  5. Jan 24, 2009 #4
    Can any1 explain this a little more clearly perhaps, I've been looking at this for atleast an hour and i'm still stuck....
     
  6. Jan 25, 2009 #5
    anyone? im still a little stuck
     
  7. Jan 25, 2009 #6
    no help?
     
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