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Homework Help: Kinematics of a point

  1. May 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Two points P and Q have position vectors in a reference frame that are given by rOP = 50ti and rOQ = 40i - 20tj meters. Find the minimum distance between P and Q and the time at which this occurs.

    2. Relevant equations

    3. The attempt at a solution
    I suppose I take the magnitude of each vector, but then what? They both share the same origin, how do i manipulate it so that they give me the PQ length?
  2. jcsd
  3. May 18, 2012 #2
    What path do each of the points move on?
  4. May 18, 2012 #3
    the x and y axis respectively
  5. May 18, 2012 #4
    What I would do here is just look at the two paths. Choose a few values of t, and see if they are getting closer together or farther apart.

    Since their paths are very simple, I don't think it would be that hard to find the point where they switch from approaching to getting more distant.

    Edit, actually I just looked at them more closely and maybe it would be easier to plug the equations into the distance formula, since you know that the triangle between them will always have its right angle on the same point.
  6. May 18, 2012 #5
    right. but i would rather there be a methodical way instead of plug and chug. can you tell me a general formula i should follow, such as take the magnitude and then subtract them ....etc. thanks
  7. May 18, 2012 #6
    I haven't attempted to solve a problem such as this before, but as a general rule I would define the equations as x1(t), y1(t), x2(t), and y2(t).

    The distance between two points is √{[x1(t) - x2(t)]2 + [y1(t) - y2(t)]2}.

    And then find the minimum value of the resulting function.
  8. May 18, 2012 #7
    I suggest using the distance formula however the coordinates will be function of time so d=f(t). You can graph d vs t and select smallest d or take the derivative d[d]/dt and set equal to zero
  9. May 18, 2012 #8
    alright i tried out the distance formula. sqrt [ (50t-40)^2+(20t)^2 ]

    I get sqrt [1900t^2-2400]

    then what? i don't think i can set it equal to anything...
  10. May 18, 2012 #9
    This is right.

    I don't believe this is right.

    sqrt [ (50t - 40)2 + (20t)2 ]
    sqrt [ 2500t2 - 4000t + 1600 + 400t2 ]
    sqrt [ 2900t2 - 4000t + 1600 ]

    This is the function for the distance between the two points, and the question asks for the minimum distance.
  11. May 18, 2012 #10


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    Staff: Mentor

    A small suggestion. If you are minimizing the distance d, then you are also minimizing d2. This may be a useful consideration when it comes to minimizing the effort required to apply calculus...
  12. May 18, 2012 #11
    ok so far i got d^2=2900t^2-4000t+1600. Now the question is to find a t such that d would be at a minimum. How do I do that? Is taking the derivative with respect to t possible? But since d depends on t, wouldn't I get 2dd'?
  13. May 18, 2012 #12
    d=Sqrt(2900t2-4000t +1600)

    Take the the derivative using chain rule, simplify, set numerator = 0, solve for t. The math will be easy
  14. May 18, 2012 #13
    so d will become zero since it's constant? and what does set numerator=0 mean? thanks
  15. May 18, 2012 #14


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    Staff: Mentor

    As I suggested, you can ignore the square root and minimize d2, that will result in a minimum d, too.

    So how do you go about finding a minimum (or a maximum for that matter) of a function f(t) with respect to t?
  16. May 18, 2012 #15
    If you are not familiar with differential Calculus, use your graphing calculator to graph the distance formula (Distance vs. time). Find the point on the graph where d is the smallest. What is the t value at that point? You can use a trace command or other features on your calculator to find the minimum d.
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