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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?

### 1 Answer

- 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.