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Acceleration word problem help!

  1. Sep 10, 2006 #1
    I'm taking highschool physics right now, and we're doing kinematics in one dimension.
    Well I've been stuck on this word problem for days and stumbled accross this
    forum on a google search.
    Thought maybe you could help. Anyway it is:
    Two soccer players start from rest, 48m apart. They run directly toward each other, both players accelerating. The first player has an acceleration whose magnitude is 0.50 m/s^2. The second player's acceleration has a magnitude of 0.30 m/s^2.
    (a) How much time passes before they collide?
    (b)At the instant they collide, how far has the first player run?
    I've tried to tackle this as any other algebra distance=rate*time problem (but with some physics equation involved)
    I attempted to get the final velocity of both runners by using the equation
    where vf=final velocity and vi=intitial.
    I used '48m' as the value for distance when I did this
    Then when I get my final velocity I'd figure out the average velocity for both runners.
    Then tackle the problem by adding their individual rate*time together to get rt(of the first guy)+rt(second guy)=48m ,Attempting to get the time they collide from that.
    BUT, apparently by plugging in 48m in the physics equation gave me a bad final velocity, therefore a bad time.
    I know the correct answer for part a is 11 seconds, but don't know how it's obtained. Please I would be greatful if anyone could help.
    I'll also try to clear this up if anyone's confused.
    Last edited: Sep 10, 2006
  2. jcsd
  3. Sep 10, 2006 #2

    Hey :)

    Both players are accelerating, so the equation for both their movement is
    first player s_1 = 1/2*a_1*t^2
    2nd player s_2 = 1/2*a_2*t^2
    with a_1 = 0.5 m/s^2 and a_2 = 0.3 m/s^2
    you know that s_1 + s_2 = 48m

    Now you sum up both players' equations and recieve:
    s_1 + s_2 = 1/2*t^2*(a_1+a_2) where t is the only onknown, so you get t!

    Then just insert the value of t in the first player's equation and you'll get s_1 which answers the 2nd question.

    have fun :)
  4. Sep 10, 2006 #3
    Oh wow!
    I feel REALLY dumb right now.
    Was the solution that simple?
    I just worked it out and got the right answer(so it is that simple :rofl: )
    I was going through every other equation, missing the easy connection with this one.
    Thanks a lot :smile:
  5. Sep 1, 2008 #4
    Re: solution

    I'm completely lost. so it would be s1+s2= 1/2t^2 (0.5 +0.30) ???
    48 = 1/2 t^2 (0.5 +0.39)
    what did i do wrong/?
    Last edited: Sep 1, 2008
  6. Sep 1, 2008 #5


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    That looks correct (other than it's 0.30, not 0.39, for one of the accelerations).

    If I were setting this up, I would have started by writing

    0 = 48 - 1/2 t^2 (0.5 +0.30),

    which results in the same thing.
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