Question 162570
2 meters east, 1 meter north.
The distance (radius) from the center of rock drop to toy boat is,
{{{D=sqrt(x^2+y^2)}}}
{{{D=sqrt(2^2+1^2)=sqrt(4+1)=sqrt(5)}}}
{{{D=2.23606}}}
So the center of the rock drop is 2.23606 meters away from the toy boat. 
The ripple is moving towards the boat at 6 cm/s. 
This is now a rate x time=distance problem.
You know the rate, the speed of the ripple.
You calculated the distance.
Now calculate the time.
First, the units have to match (can't mix cm and m).
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Change the distance to cm.
2.23606 meters= 223.6cm
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{{{R*T=D}}}
{{{T=D/R}}}
{{{T=223.6/6=37.27}}}
It will take the ripple 37.27 seconds to reach the boat.
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{{{drawing( 300, 300, -12, 12, -15, 15,grid( 1 ),circle( 0, -12, .2 ),
locate(6,6,S),
locate(-7,12,T),
locate(1,-10,U),
circle( 6, 6, .2 ),
circle( -6, 12, .2 ),
green(line(0,-12,6,6)),
green(line(6,6,-6,12)),
green(line( -6, 12, 0, -12)))}}}
M is the midpoint of the line from T to U.
{{{x[M]=(-6+0)/2=-3}}}
{{{y[M]=(12+(-12))/2=0}}}
The midpoint M has coordinates (-3,0).
The change in x from S to M is 
{{{DELTA*x=(-3-6)=-9}}}
The change in y from S to M is
{{{DELTA*y=(0-6)=-6}}}
Using only 2/3 of the change in x, you would have,
{{{(2/3)(-9)=-6}}}
Using only 2/3 of the change in y, you would have,
{{{(2/3)(-6)=-4}}}
My starting point is S at (6,6).
The point P is 2/3 of the way from S to M.
The coordinates of P are found by starting at S and adding the 2/3 change in x and y.
(6,6)+(-6,-4)=(0,2)
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P is located at (0,2)
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{{{drawing( 300, 300, -12, 12, -15, 15,grid( 1 ),circle( 0, -12, .2 ),
locate(6,6,S),
locate(1,2.5,P),
locate(-7,12,T),
locate(-3,2,M),
locate(1,-10,U),
circle( 6, 6, .2 ),
circle( -6, 12, .2 ),
circle(-3,0,.4),
circle(0,2,.4),
circle(-3,0,.2),
circle(0,2,.2),
blue(line(6,6,-3,0)),
green(line(0,-12,6,6)),
green(line(6,6,-6,12)),
green(line( -6, 12, 0, -12)))}}}