Yacht and Sea Buoy A yacht sails east from a dock with a speed of one mile per hour. Two miles northeast of the dock is a sea buoy. The event starts at t = 0⁺ with the yacht position and velocity and the position of the sea buoy indicated. When will the yacht pass closest to the sea buoy and what will be the distance from the yacht to the buoy at that time?

♦  A vector diagram is drawn (above right) with an origin placed at the dock, coordinates and unit vectors (I and J) indicated. The yacht is shown a distance from the dock at the arbitrary time, t = t^*.

The equation relating postions of the dock, yacht and buoy is:

P_yacht(t) + P_sb/y(t) = P_sb

It is a good habit to include time dependence by writing "(t)" with properties that change with time. This is a 2-D vector equation. When solved it will yield two pieces of information. When writing an equation it is helpful to do the easy parts first. Of the three vectors we need, position of the buoy is the easiest to write.

It is a good habit to include time dependence by writing "( t )" with properties that change with time. The vector equation relating postions of the dock, yacht and buoy is:

P_sb = 2 mi [cos(45°) I + sin(45°) J ]

The yacht, originally at the origin, is moving East at 1 mph for all t > 0⁺.

P_y(t) = 0 I + 0 J + 1 mph· t I

Putting these results into the equation (with time in hours):

t I + P_sb/y(t) = 2 mi [cos(45°) I + sin(45°) J ]

So everything is known except P _sb/y(t).

P_sb/y(t) = (√2 - t) I + √2 J

The distance between the yacht and buoy, D(t) is the length of this vector:

D(t) = |P_sb/y(t)| = [ (√2 - t)² + (√2)²]^1/2miles.

Next we write the above D(t) in a "completed square" form:

D(t) = [(t - √2)² + 2]^1/2miles.

So D(t) is least when t = t^* = √2 hours and the shortest distance is:

D^*( t^* = √ 2 hours) = √ 2 miles.