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When will the football players collide?

  1. Jun 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Two football players separated 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​
    . How long does it take for the players to collide?

    Given: player 1: v1=0m/s, a=2.4m/s^2
    Player 2: v1=v2=5.4m/s​
    , a=0m/s

    Conventions: right= positive, left= negative.

    Note: I did not plug in 42m for Δd because neither of the players are actually displaced 42m as they end up colliding.
    2. Relevant equations
    I used the SUVAT/5 kinematics equations, one equation for each player.

    For player 1: Δd=v1Δt + 1/2aΔt^2
    After plugging in the given values: Δd=1/2(2.4)Δt^2.
    After simplifying: Δd=1.2Δt^2

    For player 2: Δd= [(v1+v2)/2]Δt
    After plugging in the given values: Δd=[(-5.4-5.4)/2]Δt
    After simplifying: Δd=-5.4Δt
    3. The attempt at a solution
    So, I took my 2 simplified equations and set them equal to each other. My goal was to solve for Δt. I thought this would work because in math class, we found the point of intersection of two linear functions by isolating for y in both and then setting them equal to each other. We then solved for x and plugged that value back into one of the equations and solved for y.

    Both of my equations were isolated for Δd so I thought it would make sense to set them equal and solve for Δt:

    1.2Δt^2=-5.4Δt
    1.2Δt^2+5.4Δt=0

    I used the quadratic formula to solve for Δt:

    x=(-b+-√b^2-4ac)/2a
    After plugging in values and simplifying (a=1.2, b=5.4):
    x1=-5.4+√29.16/2.4
    x1=0

    x2=-5.4-√29.16/2.4
    x2=-4.5

    Both of my answers are inadmissable. Why did this happen?


    My teacher showed us the solution to this problem and she did the same thing I did except she used a different equation for player 2(Δd=v1Δt+1/2aΔt^2 -- This is the same equation used for player 1). Also, she plugged in 'x' for Δd in the equation for player 1 and 42-x for Δd in the equation for player 2. She then simplified both equations, set them equal and solved for Δt using quadratic formula.

    Why is it wrong to not plug anything in for Δd and just set the 2 equations equal to each other, since they're both isolated for Δd? Aren't we just solving for Δt at the point where Δd or player 1=Δd of player 2 (this is the point where they collide).


    In short, can someone please explain why what I did was wrong and what my teacher did was right?

    Edit: Please help! My exam is in less than 10 days and it's a grade 12 class so it's very important!!!​
     
    Last edited: Jun 13, 2016
  2. jcsd
  3. Jun 13, 2016 #2
    Let's think about the problem in terms of words and argumentation at first. I did solve this with some form of quadratic I will tell you this.

    1. do you agree that both football players consume equal time in their efforts to collide with each other (they are moving head-on towards each other)

    Their distances travelled, may vary. Same for velocities - they may vary, because player A is accelereating and the player B is moving at constant speed.
    But the time consumed should be the same for each player? we are talking about Newtonian time in this word problem of course... not einsteinian relative time.

    2. It is known that the distance will be consumed by the head-on movement of two football players.
    When the distance in-between two players has been consumed to zero metres, then the players will have collided.

    3. Even if the player A is accelerating, and player B is moving at constant speed. What do you think their combined travelling distances are? Each distance added together. You can probably eastimate at first glance, that both players are moving different distance compared to each other. But what about the total distance.


    4. otherwise small tip - you get better at these problems when you practice them. This same type of problem does come in different varieties, sometimes with not accelerating movement, sometimes with accelerating movement.
     
  4. Jun 13, 2016 #3
    1. Yes, they collide at the same point in time.
    2.Yes, I agree,
    3. 42m total because they start off 42m apart, and end up consuming all the distance in between them. So, all together, the distance travelled is 42m(if you add the distances travelled by both players)
    4.Ok, but I still don't think it should be wrong to not plug anything in for displacement, because we end up solving for the time when the displacements of the 2 players are equal(when they have collided) by setting the 2 equations equal to each other. What is the value in plugging in 'x' for displacement for player 1's equation and 42-x for displacement in player 2's equation?
     
  5. Jun 13, 2016 #4
    the constant speed guy we will call him player B

    Player A is accelerating from standstill, until some Vmax

    If player A displaces from 42m endpoint, towards some unknown point Xb

    Then what will be the endpoint for player A (you already know player A's starting point)

    I dunno actually, I just used s= (distance travelled).


    The displacements need not be equal between the players A and player B. However I will say again that for the players to have collided, the distance inbetween the players must equal zero.

    The condition of colliding in equal time between each player A and player B. The condition can still be fulfilled even if one player travelles smaller distance than the other player. As long as the total distance travelled together equals 42m

    Remember that player A is pumping-out good distance per each second.

    player B is starting from standstill position.
     
  6. Jun 13, 2016 #5
    Player A's endpoint is xb
     
  7. Jun 13, 2016 #6
    Why would you say that player A travel equal distance compared to player B.

    I will give you a hint that the player A and player B don't actually travel equal size distances. The distances differ between each player. so you made at least this one erroneous assumption.

     
  8. Jun 13, 2016 #7
    Oh, so that's why we have to sub in values for displacement- then we can sub in x for the displacement of player a and 42-x for the displacement of player b
     
  9. Jun 13, 2016 #8

    I have the correct answer for , myself, but I'm not sure whether I'm allowd to give anymore tips and hints. I don't want to get into trouble with the moderators because of revealing the correct answer.

    try to make an improved answer based upon your new knowledge and any tips you have had.

    I think you are on the correct path now. But you need to be precise when you use the Δx and the velocity vectors.

    distance = Δx
    Δx= endpoint - startpoint [I think, that was the definition]

    Also you wanted to use negative velocity for one of the players. The time for player A will be positive and time for player B will be positive.
    I recommend you choose as one of the equations

    player B
    Δx = -vt

    the race-track has known length of 42m
    there is point x0 = 0
    x1= collision point
    x2= 42m

    player B travels
    from x0--> towards x1
    x-0= -vt
    x= -vt

    now you need to figure out the player A and probably you could then use the simultenous equation (equation pair, or system of equations)
     
  10. Jun 14, 2016 #9
    ##s_a= 0.5*at^2##
    ##s_b= v*t##
    ##s_{total}= s_b+s_a##
    ## 42metres= 1.4t^2 + 5.4t##
    ##1.4t^2 +5.4t -42=0##
    ##t= \frac{-5.4±\sqrt{230.76}}{2.4}##
    t≈4.08 seconds
     
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