A boat leaves a dock at noon and heads west at a speed of 25 km/hr. Another boat heads north at 20 km/hr and reaches the same dock at 1 p.m. Where were the boats closest?

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- May 9th 2010, 03:51 PMeuclid2word problem
A boat leaves a dock at noon and heads west at a speed of 25 km/hr. Another boat heads north at 20 km/hr and reaches the same dock at 1 p.m. Where were the boats closest?

- May 9th 2010, 04:29 PMAllanCuz
First thing we need to do is draw the diagram. We get a triangle the a hypotenuse (call this D) connecting the vector of the boat going North with the vector of the boat going West.

We can find the minimum distance and thus, the points of Boats A and B.

To do this, we will use a well established formula

$\displaystyle D^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 $

Noting that the origin of the XY graph is the dock we can find a location on the x-axis for boat A and a location on the y-axis for boat B.

Boat A is moving west at a rate of 25 km/h for up to an hour. So at 1pm boat A is 25km west of the origin. Thus, the distance increases in the west (the negative direction) with time at a rate of -25t (where t is time).

Boat B is moving north at a rate of 20km/h for up to an hour. So at 12pm the boat is 20km South of the origin. Thus, the distance decreases at a rate of -20 + 20t

From here we can find points to use in our distance formula.

P1( -25t , 0 ) and P2 ( 0 , -20 + 20t )

Thus,

$\displaystyle D^2 = ( -25t - 0 )^2 + (0 - (-20 + 20t) )^2 = (25t)^2 + ( 20 - 20t)^2 $

Note that minimizing $\displaystyle D^2 $ will minimize the problem. Thus,

$\displaystyle d(t) = (25t)^2 + ( 20 - 20t)^2 $

$\displaystyle d`(t) = 50(25t) + -40(20-20t) = 0 $

Solve for t and then multiply by the given rates for boat A and boat B to find their distances