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I need help with a problem from yale physics open courseware?

  1. Dec 17, 2011 #1
    I need help with a problem from yale physics open courseware??

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

    Romeo is at x=0m at t=0s when he sees juliet at x=6m.
    a) romeo begins to run towards her at v = 5 m/s. Juliet, in turn begins to accelerate towards him at a =-2m/s^2. when and where will they cross?
    b) suppose, instead, that juliet moved away from romeo with positive acceleration a. Find a(max), the maximum acceleration for which romeo can catch up with her. for this case find the time t of their meeting. show that for smaller values of a these star-crossed lovers cross twice.

    2. Relevant equations



    3. The attempt at a solution
    i have no idea how to start this answering this question, im trying to prepare myself for my physics class im taking next semester, i just need some explanation on how to about solving the question.
    i was thinking for a) that their time would be the same when they meet each other.
     
  2. jcsd
  3. Dec 17, 2011 #2
    Re: I need help with a problem from yale physics open courseware??

    Start with a position equation for each person and then think about what will be the same when they meet.
     
  4. Dec 17, 2011 #3

    PeterO

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    Re: I need help with a problem from yale physics open courseware??

    For part (b) , and the double crossing situation, I would start by drawing a velocity - time graph for the pair. [attachment Romeo 1]

    Juliet's graph is the sloping line -she is accelerating
    Romeos line is horizontal, and starts a little later, since at his speed [5 m/s] he doesn't reach the point from where Juliet started [6 m away] until time 1.2 s.

    The consider what happens [attachment Romeo 2].
    The red area represents the "head start" Juliet made, before romeo passed her original start point.
    The green area represents the distance Romeo made up while he was running faster than the speed Juliet had attained.
    When those two area are equal - Romeo has caught - or as is suggested in this example - has run past Juliet.

    As more time passes, [attachment Romeo 3] the blue area shows the total distance Romeo gets in front of Juliet before Juliet has reached Romeo's velocity [so he gains no more distance.
    The purple area represents the distance Juliet makes up now that she is running faster than Romeo.
    Once the purple area equals the blue area she has caught - or is about to pass him ["these star-crossed lovers cross twice].

    Now had Juliet's acceleration been greater - her line would have been steeper - then the time at which the green area equaled the red area would have co-incided with the time Juliet's velocity reached Romeo's [5 m/s]

    Notice that these graphs were created in "paint" and could just as easily been sketched on paper by you - IF you got into the habit of expressing these problems like this. The velocity-time graph can be a very powerful tool.

    From my sketch I can now see that for (b) the maximum value of acceleration is 2.08333333 or 2.5/1.2. Can you see why?

    EDIT: Hmmmmm. the attachments didn't come - i will try again next post.
     
  5. Dec 17, 2011 #4

    PeterO

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    Re: I need help with a problem from yale physics open courseware??

    Hopefully attachments - note that I have deliberately not labelled axes or intercepts etc as these are merely "qualitative sketches" to show the concept.
     

    Attached Files:

  6. Dec 18, 2011 #5
    Re: I need help with a problem from yale physics open courseware??

    Thanks, i got t=1s for part a; and t=2s and t=3s for part b
     
  7. Dec 18, 2011 #6

    PeterO

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    Re: I need help with a problem from yale physics open courseware??

    The answers for part (b) should have been an acceleration value.

    With acceleration of 2, those times may well be accurate [your part (a) answer is correct but I have not checked your (b) answers], but the question was asking what acceleration value would mean Romeo only just catches up to Juliet before she continues to accelerate away.
     
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