Expert: Richard J. Raridon - 10/18/2007

Question
I have a homework problem that is frustrating me.

One plane is 730 miles east of the city traveling at 225 miles toward the city.  Another is 500 miles south and traveling at 154 miles per hour.  They are flying at the same altitude.

I figure that the square root of (730 -225t)^2   (500-154t)^2 equals the distance.

But how do I figure out how long they will see each other if visibility is 1 mile

what is the closes the planes get

And how high would visibility have to be for them to see each other for 5 minutes

Then they ask hat happens if one plane climbs to 15,000 feet higher and one descends 5,000 feet lower, how do I calculate their distance at a given time.


Thanks!  

Answer
You set [(730-225t)^2 + (500-154t)^2]^1/2 = 1 and solve for t.  Since they will both arrive at the city in about 195 minutes, t will be a little less than that.  
Just calculate how far apart they will be 5 minutes before they reach the airport.  
For different altitudes you'll need d = (x^2+y^2+z^2)^1/2