Boat Spreading

## Question:

Boat A begins from rest at point O and heads in a direction 120
degrees from the horizontal. At the same time, boat B begins from
rest at point O and heads in a direction 45 degrees from the
horizontal. If A has an acceleration of 2 feet per second per
second and B has an acceleration of 3 feet per second per second,
determine the speed of boat aA with respect to boat B at the
instant they become 800 feet apart. How long will this take?
## Answer:

I have had lots of experience with boats and never knew them to
travel extended distances at much of an angle to the horizontal
unless they were in serious trouble. But I think I know what you
mean so I won't be perverse about the choice of words.
Write the position vector for each boat, PA and PB in terms of
its x and y components where the length of the vector PA=1/2*(2)*t^{2} =t^{2} and of
PB=1/2*(3)*t^{2}
=3/2*t^{2}. The angle of the boat A vector is 120
degrees. Its x component is the cosine of 120 degrees times the
length of the vector. Its y component is the sine of 120 degrees
times the length of the vector. This gives us
PAx=-.5*t^{2} and
PAy=.866*t^{2}. The same logic for the B boat
gives us PBx=.707*3/2*t^{2} and
PBy=.707*3/2*t^{2}. The relative
distance from boat A to boat B is a vector whose x component is
PBx-PAx and whose y component is PBy-PAy. The length of that
vector is
((PBx-PAx)^{2}+(PBy-PXy)^{2})^{.5}.

The conditions is that the length of this vector be 800 feet
so
800=(.707*3/2*t^{2}+.5*t^{2})^{2}+(.707*3/2*t^{
2}-.866*t^{2})^{2})^{.5}.

Combining the coefficients of t^{2} we get
800=((1.56*t^{2})^{2}+(.1945*t^{2})^{2})^{
.5}.

Squaring both sides we get
640000=(1.56*t^{2})^{2}+(.1945*t^{2})^{
2}

Squaring the terms on the right
640000=2.43*t^{4}+0.037*t^{4}=2.467*t^{4}

t^{4}=640000/2.467=259424

t=22.6 seconds. If my arithmetic is
right this is the answer to the "How long?"
question.

To get the relative speed plug this time into the two vectors
with components VAx=-.5*2*t, VAy=.866*2*t and
VBx=.707*3/2*t, VBy=.707*3/2*t,
and calculate the magnitude of the difference.

This information is brought to you by M. Casco Associates, a
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understanding how the universe works. My name is James D. Jones.
If I can be of more help, please let me know. JDJ