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Relative motion/ Vector Question

  1. Oct 17, 2005 #1
    Please..give It A Try

    "Ship A is located 4 km north and 2.5 km east of ship B. Ship A has a velocity of 22km/h towards the south and ship B has a velociy of 40 km/h in a direction of 37(degrees) north of east.
    A)what is the velocity of A relative to B
    B)write an expression for the position of A relative to B as a function of t, where t is equal to 0.
    ***C) at what time is the separation between the ships least?
    Ive completed A and B..but i have no idea where to start for C..Can anyone help?

    wellllllllll;
    first i found the point of A relative to B=
    ...ans = [32km/h(t) + 2.5 km]i + [2.0km/h(t) + 4.0km]j
    im guessing it has something to do with a max-min problem..but im not sure where to start..i could do trial and error..but there has to be a simpler way..
     
    Last edited: Oct 17, 2005
  2. jcsd
  3. Oct 17, 2005 #2

    James R

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    If you've done (B), then you can find the distance between A and B as a function of time by taking the magnitude of the vector expression in (B).

    For example, if your answer for (B) is:

    [tex]r(t) = x(t)i + y(t)j[/tex]

    then

    [tex]|r(t)| = \sqrt{x(t)^2 + y(t)^2}[/tex]

    To find the minimum separation you need

    [tex]\frac{d(|r(t)|)}{dt} = 0[/tex]

    and solve for t. Then find |r| using the t value.

    (Hint: since [itex]|r|[/itex] is a minimum when [itex]|r|^2[/itex] is a mimimum, you can alternatively solve [itex]d(|r|^2)/dt = 0[/itex] for t, which will be easier.)

    Does that help?


    Post restored.

    --TM
     
    Last edited by a moderator: Oct 17, 2005
  4. Oct 17, 2005 #3

    Tom Mattson

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    JamesR,

    I've "soft deleted" your post. I'll restore it when suspenc3 shows some work.

    I'm cracking down on the rules. :wink:
     
  5. Oct 17, 2005 #4

    Tom Mattson

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    You're right about it being a max-min problem, but you're wrong about needing to do it by trial and error.

    How do you normally deal with max-min problems?
     
  6. Oct 17, 2005 #5
    find the quantity were trying to minimize
    find the variables involved
    find a formula relating the quantity and the variables

    This question is unlike most max min problems ive ever done
     
    Last edited: Oct 17, 2005
  7. Oct 17, 2005 #6

    Tom Mattson

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    Have you taken Calculus I yet? The reason I'm asking is that you are forgetting a crucial step, and if you have had Calc I it should be obvious what that step is. Simply setting up a relationship isn't enough.

    Do you know what I'm getting at?
     
  8. Oct 17, 2005 #7

    jtbell

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    Have you studied any calculus yet, in particular derivatives?
     
  9. Oct 17, 2005 #8
    Yes I have..forgot that ppart..oops.its been a long time since ive dealt with min problems

    so..
    d= [32km/h(t)]i + [4km/h(t)]j..Take the derivative with respect to d?

    or am i just way off
     
    Last edited: Oct 17, 2005
  10. Oct 17, 2005 #9

    Tom Mattson

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    You're getting there. But you want to take the derivative with respect to t, not d.

    I will now restore JamesR's post.
     
  11. Oct 17, 2005 #10
    yes this helps..but i still am a little fuzzy

    r(t) = 32(t)i + 4(t)j...

    0 = (radical of)[2(32t) + 2(4t)]
     
  12. Oct 17, 2005 #11

    Tom Mattson

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    No, you have to take the derivative first, and then set it equal to zero. In order to take the derivative you have to use the Chain Rule:

    [tex]\frac{d}{dx}[f(x)]^{\frac{1}{2}}=\frac{1}{2}\frac{f'(x)}{[f(x)]^{1/2}}[/tex]
     
  13. Oct 17, 2005 #12
    R1(t) = {1/2[x(t)^2 + y(t)^2]^-1/2}{x(2t) + y(2t}

    Ive been out of school for a few years..so im not sure if this is correct

    ...0=72/18t..I must have done something wrong here!
     
    Last edited: Oct 17, 2005
  14. Oct 18, 2005 #13

    Tom Mattson

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    I've started to work the problem myself, and I see that your starting point is wrong. It is close to being correct though.

    You appear to have added the position vectors for A and B, when you should have subtracted them. The position of A relative to B is [itex]\vec{r}_{AB}=\vec{r}_A-\vec{r}_B[/itex].

    I'd like for you to do the following:

    1.) Clean up your expression for [itex]\vec{r}_{AB}[/itex], and then post it here.
    2.) Obtain the expression for [itex]|\vec{r}_{AB}|[/itex], the distance between A and B, and post it here.

    Then, we can get to work on taking that derivative.
     
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