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Finding acceleration given distance?

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data
    in a women's 100-m race, accelerating uniformly, Laura takes 2.00 s and Healan 3.00 s to attain their maximum speeds, which they each maintain for the rest of the race. They cross the finish line simultaneously, both setting a world record of 10.4 s. What is the acceleration of each sprinter?

    2. Relevant equations
    vf = vi + at
    xf = xi + .5(vi + vf)t
    xf = xi + vit + .5at^2
    vf^2 = vi^2 + 2a(xf-xi)

    where vf, vi, xf and xi are final and initial velocity and horizontal position (respectively)

    3. The attempt at a solution
    xf = xi + vit + .5at^2
    100 = (0) + vit + .5a(10.4)^2
    9.62 = vi + 5.4a

    avg velocity = 100/10.4 = 9.61 m/s
    a = v/t
    a (Laura) = 9.61 / 2 = 4.8 m/s^2
    a (Healan) = 9.61 / 3 = 3.2 m/s^2

    my books answers are 5.32 m/s^2 for laura and 3.75 m/s^2 for healan. i don't know how to get there...
    I've tried using all of the above equations, but I can't get anything to work out because I have too many unknowns (vf, vi, and a). For awhile I tried using 0 as the value of vi and xi, but that didn't work out, either. I can't figure out how to use the information about the time each person reached maximum speed (or if it should be used at all).
     
    Last edited: Sep 8, 2010
  2. jcsd
  3. Sep 8, 2010 #2
    Does the formula you use really apply to the whole run? Or is there a change in the way the women run?
     
  4. Sep 8, 2010 #3
    I suppose so; I posted all the information that was given, and the formulas I listed are from the section in my book that's devoted to constant acceleration ("accelerating uniformly")... :|
     
  5. Sep 8, 2010 #4
    The formulas are correct. I wanted to hint if maybe there are some conditions on these equations, e.g. for the third there has to be a constant uniform acceleration during all time.
    Is this the case where you applied the formula?
     
  6. Sep 8, 2010 #5
    I already know that the acceleration is a constant. That's stated in the problem...
     
  7. Sep 8, 2010 #6
    I read that. But there is also something stated about maximal velocity. What does this imply for the duration of the acceleration?
     
  8. Sep 8, 2010 #7
    Well. I suppose if I knew that, I'd have a bit of an easier time with the rest of my homework ):
     
  9. Sep 8, 2010 #8
    Imagine a 100m run. And then think how much the runners still accelerate after reaching their maximal velocity.

    Sorry for not being more precise, but that would mean solving your question completly.
     
  10. Sep 8, 2010 #9
    It doesn't help much. Maybe I'm stupid. Or sleep deprived. Or just not cut out for physics (better change my major, hahha). Because all I can think is that obviously if acceleration is constant, the runner would continue to accelerate after reaching her maximum velocity. But I don't know how to approach the problem beyond that.

    Thanks for the effort, anyway. I suppose I should just give up and move on to other work that needs doing... after all, I have an essay to write and need to wake up in 4 hours... Thanks, again..
     
  11. Sep 8, 2010 #10

    rl.bhat

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    Homework Helper

    Hi talklikearobo, welcome to PF.

    distance covered by each sprinter is given by

    s = 1/2*a*t1^2 + a*t1*t2.

    Here at1 is the constant velocity and t1 + t2 = 10.4 and s = 100 m.

    Now find the acceleration of each sprinter.
     
  12. Sep 8, 2010 #11
    Don't give up so fast. :)
    The acceleration being constant means just, that it is constant as long as there is acceleration. After reaching the maximal velocity the runners do not accelerate any more, but run with constant speed.
    So maybe you should try cutting the race into two pieces. The acceleration phase and the running phase.
     
  13. Sep 8, 2010 #12

    BobG

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    Homework Helper

    Acceleration isn't constant for the entire race.

    Each runner presents two separate problems. They cover a certain distance during their acceleration. They cover a certain distance at a constant speed. The total distance of both those phases equals 100 meters.

    Your problem is that you don't know either of those distances. Instead, you know the time each phase lasted.
     
  14. Sep 8, 2010 #13
    well, I think the answers above are confusing, I try to answer the question step by step with physical reasons. let's start:

    well, the speeds of both runners at the beginning of the match is increasing with a constant rate and then when they have reached their maximum speeds their speeds remain constant for the rest of the race. that means we should split the problem into two different parts:

    Helen's velocity stays the same for all t's greater than 3. that means we can write down H(t): vmaxt + xt=3 (t>3). the maximum speed of Helen is given by another function that applies to t's less than 3 and we can write down h'(t) = at (t<3). so the maximum speed of helen when t approaches 3 equals h'(t) = 3a = vmax and its position at that time is given by h(t) = [tex] 1/2at^2 [/tex]. thus when t approaches 3 helen is at xt=3 = 4.5a meters. now if we plug these numbers into H(t) we will have : H(t) = 3at + 4.5a. at t=10.4 we have H(10.4) = 100. so we have: 3a(10.4-3) + 4.5a = 100. that means a ~ 100/26.7 ~ 3.745 m/s^2. we write 10.4-3 because we're starting from 3, not zero.

    I let you calculate laura's acceleration on your own to see if you've got the point.

    note that the position functions we're working with are not differentiable at t=3. that's a mathematical fact which you can ignore it with physical intuition (in fact they call such points removable discontinuities).
     
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