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The Slide Rules (Help me *tears*)

  1. Aug 27, 2006 #1
    Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 . Simultaneously, each puck is given a quick push and they begin to slide directly toward each other. Puck A moves with a speed of = 2.70 , and puck B moves with a speed of = 5.10 .

    What is the distance covered by puck A by the time the two pucks collide?

    Really I just need to know how to set it up becuase each time I do this problem I get a negative answer when finding time and well time can't be negative can it?

    Please help me....my brain hurts
     
  2. jcsd
  3. Aug 27, 2006 #2

    quasar987

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    What have you tried? Show your work first so we can comment on it.
     
  4. Aug 27, 2006 #3
    Well what I have been doing is that I know that X1a=X1b and the equation of x1a= x0a + V0a (t1) and x1b= x0b - V0b (t1)
    X1a= 2.70 t1
    X1b+ 18-(-5.10) t1

    2.70 t1= 18+5.10t1
    2.70t1-5.10t1=18
    -2.4t=18
    t=-7.5
    so then puck a traveled a -20.25m

    so really that what I have been doing I tweeked it to make it positive but puck a can't travel farther than 18 m so really I got stuck in a hole after that.....It has been a couple of days

    Have I been approaching it wrong?
    I setted up a list of knowns and unknowns
     
    Last edited: Aug 27, 2006
  5. Aug 27, 2006 #4

    quasar987

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    The approach is perfect. You are using the equations of kinematics under a null force with puck A at the origin and puck B at x=18.0. However, the general equation of kinematics (i.e. the "model" applicable for any more under a null force) is

    [tex]x(t)=x_0+v_0t[/tex]

    But you have been using the (wrong) equation [tex]x(t)=x_0-v_0t[/tex] for particle B. This is where your error comes from.
     
  6. Aug 28, 2006 #5
    Thank you so much! Its correct now! Just goes to show that the mastering physics dude was wrong
    Thank you
     
  7. Aug 28, 2006 #6

    andrevdh

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    Your initial equation for puck b seems fine to me, since it's x-coordinate will decrease with time. For puck a the x-coordinate will be

    [tex]x_a = 2.70t[/tex]

    and for puck b

    [tex]x_b = 18 - 5.10t[/tex]

    their x-coordinates will be the same at a time given by

    [tex]2.70t = 18 - 5.10t[/tex]
     
  8. Aug 28, 2006 #7

    quasar987

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    Well to avoid confusion, (and also because that is what the maths behind the derivation of the kinematics equation says), we take

    [tex]x(t)=x_0+v_0t[/tex]

    as the base equation, and let [itex]v_0[/itex] itself be either positive or negative wheter the motion is in the positive x-direction or negative x-direction respectively.
     
    Last edited: Aug 28, 2006
  9. Aug 28, 2006 #8

    verty

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    This question seems far easier to do as a ratio, because the distances travelled is in the same ratio as the speeds, so you can use (V_a/V_t) * S_t.
     
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