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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?
- az_lenderLv 72 months agoFavorite 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.