SOLUTION: A search plane takes off from airport at 6:00AM and travel due north at 200 miles per hours. A second plane leave that airport at 6:30AM and travel due east at 170 miles per hours.

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Question 1078342: A search plane takes off from airport at 6:00AM and travel due north at 200 miles per hours. A second plane leave that airport at 6:30AM and travel due east at 170 miles per hours. The plane carry radio with a maximum range of 500 miles. When (to nearest minutes) will these plane no longer be able to communicate with each other?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let time start when the second plane takes off.
At that point the first plane is traveling at (0,100%2B200t) where t is the time in hours from 0630AM.
The second plane is traveling at (170t,0).
The first component of the distance function is east positive, west negative.
The second component of the distance function is north positive, south negative.
We can calculate the distance between the first and second plane using the distance formula,
D%5E2=%28170t-0%29%5E2%2B%280-%28100%2B200t%29%29%5E2
D%5E2=28900t%5E2%2B40000t%5E2%2B40000t%2B10000
D%5E2=68900t%5E2%2B40000t%2B10000
Find t when D=500,
500%5E2=68900t%5E2%2B40000t%2B10000
250000=68900t%5E2%2B40000t%2B10000
68900t%5E2%2B40000t-240000=0
689t%5E2%2B400t-2400=0
Using the quadratic formula,
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-400+%2B-+sqrt%28+400%5E2-4%2A689%2A%28-2400%29+%29%29%2F%282%2A689%29+
x+=+%28-400+%2B-+40sqrt%284234%29%29%2F%282%2A689%29+
x+=+%28-200+%2B-+20sqrt%284234%29%29%2F%28689%29+
Only the positive solution makes sense here,
x+=+%28-200+%2B+20sqrt%284234%29%29%2F%28689%29+
or approximately,
x=1.5985hrs
1 hr 36 minutes from 0630AM which would be 0806AM.