# Finding closest distance

Dick
Science Advisor
Homework Helper
Good luck!

HallsofIvy
Science Advisor
Homework Helper
Because you (cleverly) chose to make A's initial position the origin of your coordinate system, A's position at t= 0 is, indeed, (0,0). Since A is moving due east (and you chose to make the positive x-axis that direction) at 8 km/h, A's position at time t hours is (8t, 0). Yep, that's what you got!

Initially, B was "5 km to the south east" so B's initial position is $(5\frac{\sqrt{2}}{2}, 5\frac{\sqrt{2}}{2}$ ($sin(45^o)= \frac{\sqrt{2}}{2})$. Since B is moving straight North at 6 km/h, B's position at time t hours must be $(5\frac{\sqrt{2}}{2}, 5\frac{\sqrt{2}}{2}+ 6t)$). If $5\frac{\sqrt{2}}{2}= 3.53$(and it is) you are correct. Glad you don't have to quit school!

Now what is the distance between those points? (Hint: since distance is always positive, minimizing distance is the same as mininizing distance squared- so you can ignore the square root in the distance formula.)

Yea I got it thanks Halls of Ivy...

I get to stay in school :D