How Do Relative Velocities Determine Collision Courses?

[Woolyabyss]

This isn't really homework its an example, But I need help understanding it to do my homework

Homework Statement

A destroyer is 500 km due west of a frigate. The destroyer is traveling at 10 km/h in a direction of 30 degrees north of east.The frigate is traveling 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?

Homework 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

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.

 

[haruspex]

"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.

It doesn't matter whether you put k on the right and have it as an inverse of time, or put it on the left and have it represent (the more obvious) collision time. The point is merely that (with the minus sign) the ratio is positive: i.e. the distance is diminishing in magnitude, so the time at which they will be at the same point is in the future.

 

[Woolyabyss]

Thanks I get it now its just that generally when I see constants of proportionality there usually put on the side of the equation where they wouldn't be inverted.

 

[Woolyabyss]

It doesn't matter whether you put k on the right and have it as an inverse of time, or put it on the left and have it represent (the more obvious) collision time. The point is merely that (with the minus sign) the ratio is positive: i.e. the distance is diminishing in magnitude, so the time at which they will be at the same point is in the future.

Sorry to bother you again but it would be fine if I said -k(Vfd) = Rfd wouldn't it?
since k would now mean time and the ratio between them would still be positive.