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Simple kinematics sprinter problem

  1. Aug 9, 2004 #1
    A sprinter reaches his max speed vmax in 2.5 sec from rest with const accel.

    He then maintains that speed and finiishes the 100 yards in the overall time of 9.6 sec. Dettermine his max speed vmax.
     
  2. jcsd
  3. Aug 9, 2004 #2
    i got a = 4.8 ft/s^2 thus vmax = 4.8*2.5sec = 12ft/s

    seens a bit fast.
     
  4. Aug 9, 2004 #3

    arildno

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    If you post the equations you've worked with, and how you attacked them, rather than posting a couple of numbers, it would be easier to spot possible mistakes you may have done.
     
  5. Aug 9, 2004 #4
    Sketch a speed time graph and calculate with the fact area = distance.
     
  6. Aug 9, 2004 #5
    sorry david, but 12 ft/sec as a max speed is a little to slow...this guy here's running 300 ft (100 yrds) in 9.6 seconds.
     
  7. Aug 9, 2004 #6
    david90, here's a way you can solve it algebraicly:

    For the first 2.5 seconds, the sprinter is accelerating....thus...we can describe that part as

    distance = (1/2)at^2, with t = 2.5 it would be = (1/2)a(2.5)^2

    For the last 7.1 seconds, we know that the speed is both constant and max....and also that the speed = a * 2.5......since distance = speed * time (which in this case is the latter 7.1)....you get...

    distance = (a * 2.5) * 7.1

    Now, just add these two equations above together, and you get the total distasnce. Set this equal to 300 ft, which is the total distance, and solve for acceleration. Once you know acceleration, you can then find the max. velocity.
     
  8. Aug 9, 2004 #7
    actually, based on your answer of a = 4.8 ft/sec^2...it appears that you must have used the above method, or a similar one...but instead of having the length be equal to 300 ft (100 yards)...looks like you had it equal to 100 ft. So anyways, i'm willing to guess that you had your method correct, but just forgot to convert your units. Just make sure in the future to keep your units consistent. :-)
     
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