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A common compare equation - please help.

  1. Sep 7, 2006 #1
    Can someone please help with this question. I know you take two equations and make them equal and then my professor told me you use the quadratic formula to get the answer but I am still no quite getting it.

    Hockey player A is standing still when player B passes him with the puck at a rate of 12m/s. Player A watches player B for 3 second then decides to catch up with him. If player A accelerate at 4 m/s^2 what will be the final distance when the two players are even? How long will it take player A to catch up with player B. I think the answer is 8.2 seconds and the distance is 132 meters.

    Can someone please give the answers by walking through the problem using the two proper euqation then the quadratic formula.
  2. jcsd
  3. Sep 7, 2006 #2
    we can help you if you show your work. have you set up the equations of motion for each player A and B separately?
  4. Sep 7, 2006 #3
    yes, but im in programming class right now and my notes are at home. I will post what I did later tonight when I get home. I just logged on an saw where you have to post some of the work you have done.

    I am not actually taking the course, but just helping a friend out who is. Of the top of my head I keep using the distance formula and getting the answer 6 seconds and a distance of 96meters. That should make it obvious what mistake I am making for a professor. I know there is a comparative way, using the two distance formulas and usuing values for the two separate players and making them equal. I also know this give you a polynomial in which the quadratic equation is used in order to get the final value (distance). Once the distance (which is a little higher then what I postered earlier) is found the second answer is easy by dividing the 12m/s and then subtracting three seconds.

    I also realize it is the 3 seconds where the skater stood there and watched the other player that throws off most students. Maybe you could give me a hint as the two formulas which are set so equal each other then i should be able to figure it out. I believe it is a variation of the d=V0*t+.5at^2. But I think you have to algebraicly change this before seting the two equal to each other. It been a few yours since I have had physics - which I loved - and i am trying to help out a friend. I took it when I got my accounting degree in 02 and am back in school working on a Master in Biofernetics - I am helping this friend out to help refresh on my physics skills as she takes the class through this semester.

    I just need some kind of hint on the initial formula to use. Shoot got to go
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