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Homework Help: Pure Kinematics problem

  1. Jul 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Two cyclists are in a race. One cyclist knows that he is slower, so he cheats: he removes the faster cyclist’s bike chain. The cheater starts from rest immediately with acceleration 2.5 m/s2. The faster cyclist has to take 3 seconds to replace her bike chain. She then follows (also from rest) with acceleration 3.0 m/s2. Assume that both cyclists accelerate smoothly and that they do not reach their maximum speeds during this race.

    What is the maximum length that the race can be (in meters) in order for the slower cyclist to win?

    The attempt at a solution

    I used [itex] x = 1/2(at^2)[/itex] and plugged in the values of the cheater cyclist into the equation.( t = 3, a = 2.5)
    Found x (came out as 11.25), plugged the same value of x into the equation with acceleration of the faster cyclist. This gave me the time it would take for the faster cyclist to reach that same position the cheater cyclist reached in 3 seconds. got t = 2.7. Plugged in t = 3+2.7 into the equation to get the position of the cheater cyclist at the time when the faster cyclist reached the position x = 11.25. Kept doing this, but it's taking me nowhere. Is there a simpler way to solve this problem?
  2. jcsd
  3. Jul 11, 2013 #2
    Can you recast the condition "the maximum length that the race can be (in meters) in order for the slower cyclist to win" mathematically?
  4. Jul 11, 2013 #3
    That would be the length just before the difference of position between them becomes zero. I don't know how to put it mathematically. I do know it involves calculus, and I can solve it if the equation is framed. But someone will have to help me/hint me on how I can get that particular equation.
  5. Jul 11, 2013 #4
    You have almost done so. The limiting case is the difference is zero. Write that down and you have it.
  6. Jul 11, 2013 #5
    Xcheater - Xfast cyclist = 0.

    Therefore, [itex]ut - 1/2at^2 = ut -1/2at^2[/itex] solving this would give me the time. How do I get the distance?
  7. Jul 11, 2013 #6
    If the second equation is a corollary of the first one, then it should have different accelerations, but it does not.

    Express ##X_{\text{cheater}}## and ##X_{\text{fast cyclist}}## correctly.
  8. Jul 11, 2013 #7
    Express how? I'm sorry. There's where the problem lies.
  9. Jul 11, 2013 #8
    Are you saying you do not know how to express displacement in uniformly accelerated motion?
  10. Jul 11, 2013 #9
    I obviously do. And yes, the second equation does have a different acceleration. But is my approach correct? To get the time from that equation, and then find the distance travelled. Or is there another way to do this problem?
  11. Jul 11, 2013 #10
    Then I suggest you do what I requested earlier: express those displacements correctly.

    Your approach is generally correct.
  12. Jul 11, 2013 #11
    Let t be the amount of time that has elapsed since the fast cyclist started out, and let t+3 be the amount of time that has elapsed since the slow cyclist started out.


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