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Acceleration Question

  1. Oct 25, 2003 #1
    I'm a physics newbie wondering if I've done this question right. I'll type the question as it is in my assignment along with my interpretation of it.

    A hockey player is standing on his skates on a frozen pond when an opposing player skates by with the puck, moving with a constant speed of 12m/s. After 3s, the first player makes up his mind to chase his opponent and starts accelerating uniformly at 4m/s^2. How long does it take him to catch the opponent?

    First, I found that after 3 seconds the opposing player (P1) will be 36m away from P2. So I calculated how long it would take P2 to travel 36m at 4m/s^2. But in that length of time P1 would have traveled an additional x number of metres. So I calculated how long it would take P2 to travel those additional x number of meters. I kept doing this until they were (roughly) at the same point.
    Strangely, the speed of P2 ended up being some unrealistic amount akin to a speeding car.

    Did I do this question correctly?
    Thanks to any who try and help. I really appreciate it. :)
  2. jcsd
  3. Oct 25, 2003 #2
    Theoretically that might work, but practically that would be a nightmare. For your problem, we are interested in when the two players are essentially occupying the same space. You started out correctly... when the second player starts after the first player, they are 36 meters apart. Now lets look at one of the kinematic equations for constant acceleration:

    x = xi + Vit + (1/2)at2

    x is the position of the hockey player relative to an origin point we select. For this case, it would be best to call the position of the second hockey player at t=0 the origin, or x=0.

    xi is the inital position of the hockey player, while Vi is the inital velocity of the hockey player. t represents time in seconds, while 'a' is the acceleration of the hockey player. We would begin by modeling both of the hockey players with this equation.

    For the first hockey player, his inital position at t=0 would be 36 meters, since that is his distance from the origin we described. We know that his inital velocity is 12 m/s, and we know that he is not accelerating, which means a = 0. Our equation for him would be

    x1 = 36 + 12t

    For the second hockey player, we are told his acceleration, and we know he starts from rest, .i.e. Vi=0, and he is initally at the origin, so his equation is

    x2 = (1/2)(4 m/s2)t2

    We are interested in the case when they occupy the same space, or

    x1 = x2

    So let's set them equal.

    (1/2)(4)t2 = 36 + 12t

    Solve for t, throw out the negative root as it doesn't make sense. You have your answer.

    This is the first time I posted homework help, so if I've given away too much forgive me. :smile:
    Last edited: Oct 25, 2003
  4. Oct 25, 2003 #3
    Thank you Antepolleo!
    That helps me greatly. I spent a very long time doing it the impractical way, but it didn't feel right :)
  5. Oct 25, 2003 #4
    I'm happy to help.

    Did I explain that alright? Do you have any questions about it?
  6. Oct 25, 2003 #5
    Actually I do.
    I'm trying to solve for t, and my algebra skills aren't what they used to be. I'm assuming by "throw out the negative root" you mean ignoring the solution of t=0.

    So, if I have this:

    Can that be simplified to:

    If so, I get t=12/2+36
  7. Oct 25, 2003 #6
    I'm sorry about that, I accidentally included the units with the acceleration and not with anything else. Try solving this:

    (1/2)(4)t2 = 36 + 12t
  8. Oct 25, 2003 #7
    (1/2)(4)t^2 = 36 + 12t
    This makes perfect sense so far, but I forget how I am supposed to factor out the t in 6t.
    Sorry, it's been a while since I've done algebra :)
  9. Oct 25, 2003 #8
    silly me, don't I need to use the quadratic formula with this equation?
  10. Oct 25, 2003 #9
    Plugging in those numbers I get 8.2 seconds.
    That makes sense. I think I'm done now :)
    Thanks again!
  11. Oct 25, 2003 #10
    Glad I could help.

    Now if only someone would answer my thread... :smile:
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