Finding closest distance

  • #26
Dick
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Good luck!
 
  • #27
HallsofIvy
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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 [itex](5\frac{\sqrt{2}}{2}, 5\frac{\sqrt{2}}{2}[/itex] ([itex]sin(45^o)= \frac{\sqrt{2}}{2})[/itex]. Since B is moving straight North at 6 km/h, B's position at time t hours must be [itex](5\frac{\sqrt{2}}{2}, 5\frac{\sqrt{2}}{2}+ 6t)[/itex]). If [itex]5\frac{\sqrt{2}}{2}= 3.53[/itex](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.)
 
  • #28
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Yea I got it thanks Halls of Ivy...

I get to stay in school :D
 

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