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Another One Dimensional Motion Problem

  1. Aug 22, 2009 #1
    1. The problem statement, all variables and given/known data


    A rock is thrown vertically upward with a speed of 12.0 s^-1 m. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 20.0 s^-1 m.

    (a) At what time will they strike each other?

    (b) At what height will the collision occur?

    (c) Answer (a) and (b) assuming that the rder is reversed: the ball is trown 1.00 s before thr ock.

    2. Relevant equations

    X = Xo + Vot + 2^-1 a t^2

    3. The attempt at a solution
    I have no idea how to even start this problem
    were the 1 subscripts are the first object that went off and the twos are the second object
    t1 = t2 - 1.00s

    plugged in

    Xo + Vo1 t1 + 2^-1 a t1^2

    Xo + Vo1 (t2 - 1.00 s) + 2^-1 a (t2 - 1.00)^2

    solved for t2 found 1 second...
     
  2. jcsd
  3. Aug 22, 2009 #2
    if i consider there velocities to be positive do i have to make there accelerations negative???
     
  4. Aug 22, 2009 #3
    do they collide at 2.152 s + 1.00 s ?????

    not sure if that's right...
     
  5. Aug 22, 2009 #4

    tiny-tim

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    Hi pointintime! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)
    If you consider distance upward to be positive, then yes, you have to make the accelerations negative. :smile:

    (and if you want us to check your result, do show the whole calculation! :wink:)
     
  6. Aug 22, 2009 #5
    how do you recomend i solve it
     
  7. Aug 22, 2009 #6

    tiny-tim

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    The method you're using looks fine :smile:

    what's worrying you about it? :confused:
     
  8. Aug 22, 2009 #7
    X1 = X2
    sense they collide

    t1 = t2
    point at which the collide

    t1 = t2 - 1.00 s

    X1 = Xo + Vo t + 2^-1 a t^2

    X1 = (12.0 s^-1 m)t1 + 2^-1 (9.80 s^-2 m)t1^2

    sense
    t1 = t2 - 1.00 s

    X1 = (12.0 s^-1 m)(t2 - 1.00 s) + 2^-1 (9.80 s^-2 m)(t2 - 1.00 s)^2
    X1 = (12. 0 s^-1 m)t2 - 12.0 m + 2^-1 (9.80 s^-2 m)(t2^2 + 1.00 s^2 - 2 t2 1.00s)
    X1 = (12. 0 s^-1 m)t2 - 12.0 m + (4.90 s^-2 m)(t2^2 + 1.00 s^2 - t2 2.00s)
    X1 = (12. 0 s^-1 m)t2 - 12.0 m + (4.90 s^-2 m)t2^2 + 4.90 m - (9.80 s^-1 m)t2
    X1 = (4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m

    sense X1 = X2

    (4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m = Xo + Vo t2 + 2^-1 a t2^2
    (4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m = (20.0 s^-1 m) t2 + 2^-1 (9.80 s^-2 m) t2^2
    (4.90 s^-2 m)t2^2 + (2.2 s^-1 m) - 7.1 m - (20.0 s^-1 m) t2 -(4.90 s^-2 m) t2^2
    (17.8 s^-1 m)t2 - 7.1 m

    what now???
    let me guess I did it wrong

    (17.8 s^-1 m)t2 - 7.1 m = 0
    (17.8 s^-1 m)t2 = 7.1 m
    t2 = .3989 s

    what does that mean????
     
  9. Aug 22, 2009 #8
    can someone check my work i probably did it wrong and if i didn't i don't know what to do from there
     
  10. Aug 22, 2009 #9

    tiny-tim

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    Well, it's bit difficult to read (please use the X2 tag in future), but the method looks ok,

    except (i thought you agreed on this) the acceleration needs to be negative. :rolleyes:
     
  11. Aug 22, 2009 #10
    It does have to be negative???
    If I consider it to be positive does it miss anything up
     
  12. Aug 22, 2009 #11

    tiny-tim

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    Yeees! …

    if your 12 and 20 are positive, then your 9.8 must be negative (or vice versa). :wink:

    what's up for velocity must be up for acceleration!​
     
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