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Kinematics Football Problem -- 2 football players colliding...

  1. Feb 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Two football players seperated by 42m run directly toward each other. Football player 1 starts from rest and accelerates at 2.4m/s^2, and football player 2 moves uniformly at 5.4m/s.
    a) How long does it take for the two players to collide?
    b) how far does each player move?
    c) How fast is football player 1 moving when the players collide?

    For part a- player 1: displacement=42m, a=2.4m/s^2, vi=0m/s.
    Player 2: vi=vf=5.4m/s, displacement=42m
    .

    2. Relevant equations
    I'm only at part a: For player 1, I used the eqn. displacement=vi(t)+1/2a(t)^2
    For player 2, I used displacement= (vf+vi/2)t.

    3. The attempt at a solution
    I plugged in all known values for both players. I tried this twice- the first time, I plugged in the displacement for each player. The second time, I did not. But, for both attempts, I set the 2 equations equal to each other, moved all variables over to one side of the equals sign, setting the equation equal to 0. Then, I used the quadratic formula and solved for time. I got the same answer both times, 4.5seconds. But, the correct answer is 4.1s. I have the solution for this problem, but it's very confusing. What did I do wrong?

    I would type up all of my work, but it's over a page long and would look very confusing/messy if I tried to type it


    Here's what I did:

    player 1: displacement=o(t)+1/2(2.4m/s^2)t^2

    displacement=1.2m/s t^2


    Player2: d=5.4m/s(t)

    Then, I set the 2 equations equal to each other:

    5.4m/s^2
    (t)=1.2m/s(t)^2

    I moved all terms over to one side of the equals sign: 1.2m/s
    (t)^2-5.4m/s^2(t)=0
    Then, I used the quadratic formula to solve for t.

    x=5.4+√29.16/2.4

    x1=4.5 seconds. x2=0seconds

    Therefore, time=4.5seconds. But, the correct answer is 4.1seconds.
     
    Last edited by a moderator: Feb 20, 2016
  2. jcsd
  3. Feb 20, 2016 #2

    gneill

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    Staff: Mentor

    We can't critique what you don't post. So you'll have to show your work in order to get help with it.

    One suggestion: Make sure that your equations of motion for each player are with respect to the same coordinate system. They should give both give positions versus time that are a true reflection of their locations in that system (Do both players start at x = 0? Are both player's velocities positive?).

    Out of curiosity, what were you trying to accomplish with the odd formatting?
     
  4. Feb 20, 2016 #3
    If they were with respect to the same coordinate system, one player would start at 0m and one would start at 42 m, right?
    I didn't format it like that, it just turned out that way and I have no idea why.
     
  5. Feb 20, 2016 #4

    gneill

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    Right.
    I've removed the BBcode tags from the OP that were causing the odd formatting.
     
  6. Feb 20, 2016 #5
    Here's hwta
    I added my solution in the original post, but the formatting is weird again
     
  7. Feb 20, 2016 #6

    gneill

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    You must be cutting and pasting formatted text from some document that includes formatting tags. Try pasting as "unformatted text" into a text document in a simple editor (like Notepad) first.
     
  8. Feb 20, 2016 #7
    It's fixed now. Can you check my solution?
     
  9. Feb 20, 2016 #8
    So anytime you have a question with more than one frame of motion, like this, you write all of your known values with reference to the same coordinate system?
     
  10. Feb 20, 2016 #9

    gneill

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    It does not incorporate my suggestion in post #2; You have the second player starting from an initial displacement of zero, and the two players are not moving towards each other.
     
  11. Feb 20, 2016 #10

    gneill

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    Yes. Otherwise how can you say where one is with respect to the other?

    In this case there is only one frame of reference, that of the football field. It's up to you to lay down a coordinate system in the frame of reference and express both player's motions in that coordinate system.
     
  12. Feb 20, 2016 #11
    makes sense, I'll try the problem again
     
  13. Feb 20, 2016 #12

    SteamKing

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    Since Player 2 moves at a constant velocity of 5.4 m/s throughout his run, then the distance he travels is simply d = v * t.

    Writing 5.4 m/s2 t suggests he is accelerating, which is not the case here, and the units are not correct in any event.
    (What units are used to measure distance?)
     
  14. Feb 20, 2016 #13
    But when you set 2 equations equal to each other, don't you have to have 2 variables in each equation that you don't plug anything in for? That's what I remember from math class. So, shouldn't we not plug anything in for displacement?
     
  15. Feb 20, 2016 #14

    SteamKing

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    It's not clear what you mean here.

    For this problem, the two players are running toward one another after being separated initially by a distance of 42 meters. Each second the players are running, there is a different distance being shaved off this initial separation. The amount being shaved off is related by the time spent running, which elapses the same for each player. After a certain amount of time, the players are going to collide with one another, as the distance between them will reach zero.

    It always helps to make a sketch of a problem to clarify relationships and what is changing in time.
     
  16. Feb 20, 2016 #15
    For example, if you set the equation of a parabola and a line equal to each other, you can find the point of intersection. You would get an equation like y=3x^2 and y=4x+2, where you have 2 unknowns in each quation, y and x, and you plug in one equation for y in the other equation. so, shouldn't our 2 equations have both d and t as the 2 variables without any values plugged in? And then, we can sub one equation in for d in the other equation, if that makes sense
     
  17. Feb 20, 2016 #16

    gneill

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    That's exactly what you're doing here. You have two equations for position versus time. You take the expression for one position and substitute it for the position variable in the other equation.
     
  18. Feb 20, 2016 #17
    But, we plugged something in for d, which is like the variable y in those equations
     
  19. Feb 20, 2016 #18

    gneill

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    I don't understand your confusion. You have two variables, d and t, in each equation. How you go about solving the two equations in two unknowns is up to you. You can do the easy thing and equate the two positions (d's) and solve for t first, or substitute t from one into the other and solve for d first. Either way its a standard algebra problem at that point.
     
  20. Feb 20, 2016 #19

    I tried this problem again, plugging in +42m for d for player 1 and -42m for d for player 2 and I got the same answer as before
     
  21. Feb 20, 2016 #20

    gneill

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    Why did you plug in values for d? It's an unknown. The only positions you know are the initial positions, neither of which are "d".
     
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