Relative motion involving vectors(kinematics)

[Jharr94]

Homework Statement

Ship A is located 3.5 km north and 2.8 km east of ship B. Ship A has a velocity of 22 km/h toward the south and ship B has a velocity of 40 km/h in a direction 37° north of east.

At what time is the separation between the ships least?

Homework Equations

V_ab=V_b-V_a
Where V_ab is the velocity of a with respect to b and V_a or V_b is the velocity of that ship.

The Attempt at a Solution

I'm thinking this part of the problem involves relative motion as well as kinematics, however I can't find out how to get the LEAST distance, the only thing I can think is when the positions are equal, which they should never be however,

I've found the position vector of a with respect to b as:

(2.8-31.95t)i+(3.5-46.07t)j=r_fab (r_fab is final position of a with respect to b)

from this I'm stuck as to where to go, I don't know how to find the LEAST distance between the two, I tried setting the position equations equal to each other( r_b=r_ab) but I can't seem to find answer because there are two different times for the x component and the y component.

Help please, thanks.

 

[HallsofIvy]

If the relative position vector is (a+ bt)i+ (c+ dt)j, then the distance between them is $\sqrt{(a+bt)^2+ (c+ dt)^2}$. Do you know how to minimize that?