## 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?

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)*t2 =t2 and of PB=1/2*(3)*t2 =3/2*t2. 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*t2 and PAy=.866*t2. The same logic for the B boat gives us PBx=.707*3/2*t2 and PBy=.707*3/2*t2. 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*t2+.5*t2)2+(.707*3/2*t 2-.866*t2)2).5.

Combining the coefficients of t2 we get 800=((1.56*t2)2+(.1945*t2)2) .5.

Squaring both sides we get 640000=(1.56*t2)2+(.1945*t2) 2

Squaring the terms on the right 640000=2.43*t4+0.037*t4=2.467*t4

t4=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.

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