Nice question ,
Given stuff
A :{x(t)=3-t , y(t)=2t-4}
B :{x(t)=4-3t , y(t)=3-2t}
Answer ,
1. The initial position of the boats can be find out by simply substituting the values of t=0 which yield there initial position
for color(blue){A(3,-4), B(4,3)}.
2. To find the velocity we know that the velocity is the first derivative of the position vector ,
for A,
x(t)=3-t
x'(t)=3
y(t)=2t-4
y'(t)=2
For B similarly we get
x'(t)=-3
y'(t)=-2
for the resultant we know that the formula for resultant vector is
vec ("resultant")=sqrt(x^2+y^2-2xycostheta)
we get for A we have we get sqrt(3^2+2^2)=sqrt(13)
similarly for B we get sqrt(13) but with different direction .
3. for finding out the angle between them we need to make its trajectory a vector , we know its initial coordinates, now we find its coordinates after 1 second then we use it as a vector ,
position vector after 1 sec of A boat and B boat are color(red)((2,-2), color(red)((1,1) respectively ,so we make there vector which is (-hati+2hatj) and (-3hati-2hatj) after that we have formula
cos theta=(hatp*hatq)/(|p|*|q|)
to get the angle between the vectors we get theta=cos^(-1)(-1/sqrt(65))
4.