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.#
