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Kinematics for hockey player

  1. Jun 7, 2008 #1
    1. A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 15 m/s, skates by with the puck. After 3.5 s, the first player makes up his mind to chase his opponent.

    (a) If he accelerates uniformly at 4.0 m/s^2, how long does it take him to catch his opponent? (Assume the player with the puck remains in motion at constant speed.)

    (b) How far has he traveled in this time?



    I've tried to solve it by coming up with an equation for the position of each player, and set them equal to each other. Equation for the player who is chasing: X = v*t

    I'm not sure what formula to manipulate in order to come up with an equation for the position of the player being chased.

    Thanks in advance.
     
  2. jcsd
  3. Jun 7, 2008 #2

    Kurdt

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    The equation you have for the chasing player only involves speed but he is accelerating. That would be better suited for the player being chased. Are you familiar with the kinematic equations?
     
  4. Jun 7, 2008 #3
    My bad, I wrote it opposite. X = v*t is for opponents position
     
  5. Jun 7, 2008 #4

    Kurdt

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    Remember the opponent will have traveled some distance in the thinking time as well. Is there any equations in the link I gave that you can see helping you for the chasing player?
     
  6. Jun 7, 2008 #5
    X = Xi + Vi(t) + 1/2(a)(t^2)

    Xi = 52.5
    Vi = 0
    a = 4

    Chaser --> X = 52.5 + 1/2(4)(t^2)
    Opponent --> X = v*t

    Lost at this point
     
  7. Jun 7, 2008 #6

    Kurdt

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    It would be best to add the distance the opponent has traveled onto the opponents equation and just have the acceleration in the chasers equation. From there you want to make them equal and then you will have a quadratic equation in t to solve.
     
  8. Jun 8, 2008 #7
    The initial separation is (15*3.5)=52.5 m
    Let they meet after "t" seconds
    15t=0.5*4*(t*t)+52.5
    solve for "t"
     
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