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2 bodies travelling towards each other at uniform acceleration

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data
    P and Q are points 162m apart. A body leaves P with initial speed 5m/s and travels towards Q with uniform acceleration 3m/s^2. At the same instant another body leaves Q and travels towards P with initial speed of 7m/s and uniform acceleration 2m/a^2.

    After how many seconds do they meet and what then is the speed of each body?


    2. Relevant equations
    Are there any equations?


    3. The attempt at a solution
    I drew out drew out a speed-time graph with both objects but it gave me nothing..but then why should it
    I'm absolutely clueless now
    I've seen examples of similar questions online but they don't have acceleration

    Please help :(
     
  2. jcsd
  3. Oct 29, 2012 #2

    gneill

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    Staff: Mentor

    Hi Nimrod 7, Welcome to Physics Forums.

    What formulas do you know concerning motion with acceleration?

    One approach is to choose a frame of reference (coordinate axes) with its origin at some convenient location and write the equations of motion for the two moving bodies against that set of axes. They meet when they both have the same coordinates.
     
  4. Oct 29, 2012 #3
    I've been usng these forumulas without a problem for a few days
    a = v-u/t
    v = u + a t
    s = u t + 0.5 a t^2
    v^2 = u^2 + 2 a s
    but none of them seems adequate for this problem

    This is a new type of questions in my book and it has no examples of them..

    I don't really understand frame of reference, could you explain?
    I have no teacher, I'm doing Applied Mathematics on my own :/
     
  5. Oct 29, 2012 #4

    gneill

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    Staff: Mentor

    Okay. No problem. Since the accelerations involved are both constant values, you won't be required to compute an acceleration, so you won't require the first of your equations.

    You will be needing to compute some velocities, so hang onto the second equation.

    You will also have to calculate distances, so you'll be needing the third of your equations. I would add one more parameter to it, though, namely the initial position:

    s = so + u t + 0.5 a t^2

    That way you can start each object off at some initial location.
    A frame of reference is the 'view' that some theoretical observer has of the action taking place. Usually this 'observer' is at some fixed location, and he erects coordinate axes at his position which he uses to measure the goings on around him. In this case, for example, you might place the origin of the axes at the location of the object P, with the x-axis extending from there through the object Q. Thus the launched objects will travel along the x-axis (the one from P moving in the + direction towards Q, while the one from P moves in the - direction towards the origin at P).
     
  6. Oct 29, 2012 #5
    So this is what I did..
    http://www.dumpyourphoto.com/files5/147040/thumbnail/bWwyF36.jpg
    The answers match to the answer at the back of the book and I understand most of it except...
    Why S1 + S2 = 162
    Why do they add up to 162?
     
  7. Oct 29, 2012 #6

    gneill

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    Staff: Mentor

    Because in your calculations the distances S1 and S2 represent the distances of each object from their starting positions, and you are interested in the moment in time when they meet. Since they started out separated by 162m, S1 and S2 must have covered a total of 162m when they meet.

    An alternative formulation would set the position of the first object at

    x1 = u1*t + (1/2) a1*t2

    and the second, given its starting position 162m away, at

    x2 = 162 - u2*t - (1/2)a2*t2

    Then you find the time t where x1=x2.
     
  8. Oct 29, 2012 #7
    Oh I get it now. I just did a similar question there, it's fine now :)
    Thanks for your help, Neill
    I really appreciate it
     
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