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How fast is the distance changing? (FODE, I think)

  1. Jul 30, 2010 #1
    1. The problem statement, all variables and given/known data
    2 cars start from the same point. Car A travels a constant 50 mph due west. Car B travels a constant 27 mph due south. After 3 hours, how fast is the distance changing between them?


    2. Relevant equations



    3. The attempt at a solution

    I saw this problem online have been stumped trying to solve it. To start I drew a right triangle according to the problem. I have car A travelling west and car B travelling south. I labeled the hypotenuse side C. I tried to write an equation relating the three sides and (using phythagoras) I said that:

    [tex]C=\sqrt{A^2+B^2}[/tex]

    My thought was to take the derivative of this equation and then find it's value at 3 hours. I don't think this is correct because I would be taking the derivative with respect to time (I think) and I am not sure how to handle both A and B.

    I would like a small hint in the correct direction. I am anxious to solve this one myself, if possible please hold off on presenting the solution.

    Thanks for taking the time to read and help.
     
  2. jcsd
  3. Jul 30, 2010 #2

    Mark44

    Staff: Mentor

    Instead of using A and B, you need expressions that represent the distance each car travels, as a function of time.

    dA = 50t
    dB = 27t

    Now, what expression gives you the distance between the two cars as a function of time t? (What you had is a step in the right direction.)
     
  4. Jul 30, 2010 #3

    Mark44

    Staff: Mentor

    BTW, what does FODE stand for?
     
  5. Jul 30, 2010 #4
    Sorry about the abbreviation. In my DiffEq book, they use FODE to mean First Order ordinary Differential Equation.

    This problem does not come from my DiffEq book, but it did remind me of a similar problem in that book that also stumped me.

    Thank you for the hint. As always your posts lend insight. I will work on this a bit more and then reply.
     
  6. Jul 30, 2010 #5
    Hmm, ok. Using your suggestions here is what I did:

    I re-wrote my equations (Thanks, organization is crucial) as:

    [tex]C(t)=\sqrt{(50t)^2+(27t)^2}[/tex]

    I then simplified as follows:

    [tex]C(t)=\sqrt{t^2(50^2+27^2)}[/tex]

    ...then:

    [tex]C(t)=\sqrt{t^2(3229)}[/tex]

    ...and finally:

    [tex]C(t)=t{\sqrt{3229}}[/tex]

    I stopped here, because it implies that the distance grows at a constant rate. I originally thought this makes sense, since the cars are travelling at constant velocities . I then thought that this couldn't be correct because I was under the impression this problem involved changing rates (which is why I thought it was a FODE).

    I assumed it would be a DiffEq problem because it was similar to a previous DiffEq problem I have encountered.

    Does this look correct?
     
  7. Jul 30, 2010 #6

    Mark44

    Staff: Mentor

    Looks fine to me. As a sanity check, notice that C(0) = 0, C(1) = sqrt(3229), and C(2) = 2 sqrt(3229). After 2 hours, the cars are twice as far apart as they were after 1 hour.
     
  8. Jul 30, 2010 #7
    Thanks. This makes sense to me also. I would like to point out something about this problem. I am not sure how relevant this is because the "correct" answer could have come from anywhere. I found this problem on another forum and could not solve it. The original poster came up with the same answer as I now have (about 57 mph) and they stated it was not correct. The poster said the "correct" answer is about 68mph and that they (the poster) did not understand how to get that answer.

    Again, I have no idea where the "correct" answer comes from and I have no idea whether or not the original poster posted the problem correctly.

    I wanted to add that bit of information to see if it struck a chord.

    Thanks for the help! Unless that last bit changes anything, I am satisfied with the answer.
     
  9. Jul 31, 2010 #8

    Mark44

    Staff: Mentor

    I can't find anything wrong with what you did, so an answer of about 57 mi/hr looks good to me. I don't see how the other person came up with an answer of 68.
     
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