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Please check my acceleration hockey answer.

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data

    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 12 m/s. After
    3.0 s, the first player makes up his mind to chase his opponent and starts
    accelerating uniformly at 3.8 m/s^2.
    (a) How long does it take him to catch his opponent?
    (b) How far does he travel before he catches up with his opponent?


    2. Relevant equations



    3. The attempt at a solution

    We know the initial velocities and accelerations of the two players,
    a1 = 4 m/s2, a2 = 0
    v1i = 0, v2i = 12 m/s
    x1i = x2i = 0

    The position of player #2 is given by
    x2 = x2i + v2i t + (1/2) a2 t2
    x2 = (12 m/s) t

    Be careful with the time. We must account for player #1's wait of 3 s. With this accounted for, we can calculate the position of player #1 from
    x1 = x1i + v1i (t - 3 s) + (1/2) a1 (t - 3 s)2

    Of course, this equation only makes sense for t > 3 s.
    x1 = 0 + 0 + (1/2) (3.8 m/s2) (t - 3 s)2 = (2 m/s2) (t2 - 6 s t + 9 s2)

    Now we set x1 = x2 and solve for the time t.
    x1 = x2
    (1.9 m/s2) (t2 - 6 s t + 9 s2) = (12 m/s) t

    We can either keep the units in explicitly or ensure that we have consistent units and simply write

    1.9 (t2 - 6t + 9) = 12t
    1.9 t2 - 12 t + 18 = 12 t
    1.9 t2 - 24 t + 18 = 0

    There are two solutions to this quadratic equation,
    t1 = 11.8 s, and t2 = 0.8 s

    However, the equation for the position of player #1 is not valid for t2 < 3 s, so we keep only tx,

    t = 11.8 s

    (b) How far has the first player traveled in this time?
    Now where is player #2 (and, therefore, player #1 as well) at this time?
    x2 = x2i + v2i t + (1/2) a2 t2
    x2 = (12 m/s) t
    x2 = (12 m/s) (11.8 s)
    x2 = 141.6 m

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 15, 2008 #2

    LowlyPion

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    It's really simpler if you start your time at the point that the second player starts to accelerate.

    The player at rest is at x=0 at t=0.
    At that moment player 1 is 36 meters ahead and moving at 12 m/s.

    That makes it that when x for both is equal that will be at

    1/2*3.8*t2 = 3*12 + 12*t

    1.9*t2 - 12*t - 36 = 0

    You know the quadratic formula from here. And either formula should yield the distance.
     
  4. Nov 15, 2008 #3
    i get 2.21 if i use the quadratic equation for part b..
     
  5. Nov 15, 2008 #4

    LowlyPion

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    When you have solved for t, you only have to plug that into x = 1/2*a*t2
     
  6. Nov 15, 2008 #5
    o ok, so my answer of 141.6 was right
     
  7. Nov 15, 2008 #6

    LowlyPion

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    It's not what I get because you made a rounding simplification in yout equations.

    This math is not correct:
    If you solve the quadratic for what I supplied you should get something like 8.53 sec.

    Your time is based on the first skater's start time and is 3 sec longer than the correct answer in any event.
     
  8. Nov 15, 2008 #7
    hmm.. when i plug
    1.9 t2 - 24 t + 18 = 0

    into the quad. eq i get 11.8308 or 0.800
     
  9. Nov 15, 2008 #8

    LowlyPion

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    Yes and that is wrong from the statement of the problem.

    Your time = 0 in your equations is 3 seconds before the skater actually starts. You must subtract 3 seconds from that.

    But that aside you have not correctly reduced your own equation. Your quadratic is incorrect from the rounding that you introduced.
     
  10. Nov 15, 2008 #9
    ok so then 11.8 - 3 = 8.8, and then that also solves the problem with the sig figs since the smallest number in my quadratic equation (1.9) has only 2 sig figs.
     
  11. Nov 15, 2008 #10

    LowlyPion

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    Why don't you try solving your own equation correctly.
    You have failed to correct your math error. You came up with :
    1.9*t2 - 24*t + 18 = 0

    If you had reduced your math correctly it would have been:
    1.9*t2 - 23.4*t + 17.1 = 0

    This yields the right answer after subtracting 3.
    It bears a remarkable resemblance to my method I might note.
     
  12. Nov 15, 2008 #11
    ok when i plug in 1.9*t2 - 23.4*t + 17.1 = 0

    i get 11.83-3 =8.83--> sig figs = 8.8 seconds
     
  13. Nov 15, 2008 #12

    LowlyPion

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    Your answer is wrong still. You didn't solve the quadratic for that equation.
     
  14. Nov 15, 2008 #13
    really!? what exactly are you getting?
     
  15. Nov 15, 2008 #14

    LowlyPion

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