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Diffrential calculus; distance problem

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data
    Two cars approach an intersection at the same time. The red car is 300m east of the intersection and traveling at a speed of 60km/h. The blue car is 450m north of the intersection and traveling at a speed of 75km/h. When are the two cars the closest?


    2. Relevant equations
    c^2 = a^2 + b^2


    3. The attempt at a solution

    I converted the speeds into m/s first then used it in a pythagorean theorem equation and found the derivative.
    75km/h = 20.8m/s 60km/h = 16.7m/s
    c^2 = (20.8t - 450)^2 + (16.7t - 300)^2
    and when i do the derivative i get 20.2. But I was just wondering, does it make a difference whether i do 20.8t - 450 or 450 - 20.8t??
     
  2. jcsd
  3. Dec 8, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Here it doesn't matter because of the squares. 20.8t- 450 is the negative of 450- 20.8t but squaring removes the difference.

    But you get "20.2" for what? This problem asks when the two cars will be closest. That requires a time answer. If you mean "The two cars will be closest 20.2 seconds after the starting time (when they are at the given positions)", you must include the "seconds".
     
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