This isn't really homework its an example, But I need help understanding it to do my homework 1. The problem statement, all variables and given/known data A destroyer is 500 km due west of a frigate. The destroyer is travelling at 10 km/h in a direction of 30 degrees north of east.The frigate is travelling at 5(2)^(1/2) in a NW direction. (i) Find the velocity of the frigate relative to the destroyer. (ii) Show that they are on a collision course. (iii) When will they collide? 2. Relevant equations Vab = Va -Vb (V is a vector I don't know how to write vector notation on a computer) d is the destroyer f is the frigate 3. The attempt at a solution (solution of example as in the book) (i)Vd = 10cos30i + 10sin30j = 8.66i + 5j Vf = -5(2)^(1/2)cos45i + 5(2)^(1/2)sin45j = -5i +5j Vfd = (-5i +5j) - (8.66i +5j) =-13.66i (ii) Position of frigate relative to the destroyer is at 500i km we write this as Rfd = 500i km the velocity of the frigate relative to the destroyer is Vfr = -13.66i km/h (I think this is an error and it should Vfd) since Vfd = -k(Rfd) where k is a positive constant , they must be on a collision course. (iii) the time of the collision is given by relative distance/ relative speed 500/13.66 = 36 hours and 36 minutes later. I understand (i) completely my main problem lies with (ii) where it says "since Vfd = -k(Rfd) where k is a positive constant , they must be on a collision course." The only way I can see they could make them both equal is if k were the inverse of time and if it is then why not just put k on the left and let it equal to time. I also don't get why the negative sign is needed.If anybody could help explain this one line to me it would be greatly appreciated.