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Two Knights Collide

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    In an historical movie, two knights on horseback start from rest 80.5 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.377 m/s2, while Sir Alfred's has a magnitude of 0.364 m/s2. Relative to Sir George's starting point, where do the knights collide?


    2. Relevant equations
    a kinematic equation?
    V^22nd-(Vo^2,1st + 2a1st *x1st) / 2a2nd

    3. The attempt at a solution
    a1st=.377
    vo1st=0 m/s
    x1st=80.5

    a2nd=-.364
    Vo2nd= 0m/s
    x2nd=80.5

    im so lost this is my first physics class ever and this is hw from chapter 2...
     
  2. jcsd
  3. Feb 1, 2009 #2

    Kurdt

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    Welcome to PF!

    The first thing to do is choose and appropriate kinematic equation by looking at the variables you have and the one that you want.

    https://www.physicsforums.com/showpost.php?p=905663&postcount=2

    Secondly set up the equations correctly remembering that St George's position is the reference point, and thirdly solve them.
     
  4. Feb 1, 2009 #3
    i do not know what to do with two accelerations? also am i have had one class and this is my homework, the professor just posted the equations on the board. i'm teaching myself from square one...

    so wold i use
    velocity & displacement:
    v^2 = v_0^2 + 2 a Delta x

    also wouldn't sir alfreds be negative cause he is moving backwards(Left) on the x plane
     
    Last edited: Feb 1, 2009
  5. Feb 1, 2009 #4

    Kurdt

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    You will have two equations one for each knight. Then you can solve them simultaneously.
     
  6. Feb 1, 2009 #5
    o-i-c

    so
    (0)^2+2(.377)(80.5)=60.69
    (0_^2+2(-.364)(80.5)=-58.6
    neither are correct,i apologize i hate not even having common sense,but im so lost in this stuff..
     
  7. Feb 1, 2009 #6

    Kurdt

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    First off you're trying to find the distance they travel when they hit each other so you can't use the distance between them. You may want to find a different kinematic equation to use.
     
  8. Feb 1, 2009 #7
    x = x_0 + v_0 t + (1/2) a t^2 ?
     
  9. Feb 1, 2009 #8

    Kurdt

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    Yes thats the one. Now remember the reference position is for one of the knights so be careful setting up the other knights equation.
     
  10. Feb 1, 2009 #9
    so i use that formula twice and do what with the two numbers? subtract or something
    what is X_0 and the t?
    do i convert the .377 and .364 into seconds
     
  11. Feb 1, 2009 #10

    Kurdt

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    You need to set up the equation for each knight. That is put in the variables yu know for each particular knight. One of them starts at the origin so what will x_0 be then? The other starts 80.5 meters away from the origin so what will x_0 for that knight be and what sign will his acceleration have? Don't worry about the uknonws for now. If you need a bit more help understanding why you're doing this have a look for simultaneous equations.

    http://www.tech.plym.ac.uk/maths/resources/PDFLaTeX/simultaneous.pdf
     
  12. Feb 2, 2009 #11
    80.5+V_0t+(1/2)(.377)T^2

    AND

    -80.5+V_0t(1/2)(-.364)t^2

    but what do i do with the t and V_0t....
    i really think this class is above me...
     
  13. Feb 2, 2009 #12

    Kurdt

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    Don't give up! It should become clearer once you've solved one example and try another.

    V0 is the initial speed of each knight, what are their initial speeds.

    One more thing the first knight starts at the origin (i.e. x0 = 0), and the other is 80.5 metres away not -80.5. once you have that sorted you can solve the equations by eliminating one of the 2 unknowns.
     
  14. Feb 2, 2009 #13
    0+V_0t+(1/2)(.377)T^2

    AND

    80.5+V_0t(1/2)(-.364)t^2
     
  15. Feb 2, 2009 #14

    Kurdt

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    That looks ok except there is a + sign missing from the second one. So their initial speeds are zero so you can get rid of the terms with V0 in there. Then you'll be ready to solve simultaneously.
     
  16. Feb 2, 2009 #15
    man i need more help im not asking for answers but something has to give...i have spent over four hours on this problem and nothing i have not examples,no hard copy practice its just lecture and online hw this is bs...

    idk what i'm doing i'm questioning my basic algebraic skills now.
    ihave:

    0+T+(1/2)(.377)T^2::: T^2=.1885
    80.5+T+(1/2)(-.364)T^2:::T^2=-.182
     
  17. Feb 2, 2009 #16

    Kurdt

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    Now you have:

    [tex]s = \frac{1}{2} a_1t^2 [/tex]
    [tex]s = 80.5 - \frac{1}{2}a_2t^2[/tex]

    Now eliminate the variable you don't need (i.e. the time). Check out the link I gave you about simultaneous equations for more info on that.
     
  18. Feb 2, 2009 #17
    im sorry but i dont what to do, all this is just confusing me more.i need answers
    i have forced my self to go this far. if i plug it all in it will make sense
     
  19. Feb 2, 2009 #18

    Kurdt

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    I'm afraid we can't give out answers only hints. I've told you what you need to do, so do it. You want the distance s and you want to get rid of t. Make t the subject for one of the equations and substitute it into the other, then solve for s.
     
  20. Feb 2, 2009 #19
    1/2(.377)t^2=80.5-1/2(.364)t^2
     
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