Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Anonymous asked in Science & MathematicsMathematics · 2 months ago

calc. problem help?

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 15 km/h. Another boat has been heading due east at 20 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together?

1 Answer

  • 2 months ago
    Favorite Answer

    Let's put the dock at the origin of a coordinate system, and we'll say t = 0 at 2 p.m.  The first boat's motion can be described by

    y(t) = -15t

    and the second boat's motion by 

    x(t) = 20t - 20.

    The distance between the boats obeys

    r^2 = x^2 + y^2, or

    2r dr/dt = 2x dx/dt + 2y dy/dt,

    so dr/dt will be zero only when

    x dx/dt + y dy/dt = 0, or

    20x - 15y = 0,

    which is to say,

    y/x = 20/15.

    Therefore, at the time of interest, you have

    (20 - 20t)/(15t) = 20/15 =>

    400 - 400t = 225t =>

    400 = 625t =>

    t = 400/625 (of an hour), which is 38.4 minutes.

    To see that this makes some sense, note that at this time,

    y would be -9.6 km and x would be -7.2 km, 

    so the distance between them would be 12 km,

    which is certainly smaller than the distance at either 2:00 or 3:00.

Still have questions? Get your answers by asking now.