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Old 24th Jul 2004, 15:36
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Genghis the Engineer
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Join Date: Feb 2000
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Bit of high school algebra will do it.

If we assume that t=time in minutes, s=distance in nautical miles.

At t=0 (actually 10am), Aircraft A is at distance 0, and aircraft B is at distance 300 (the other airfield).

So, position of A is (100/60)*t
And position of B is 300 - (150/60)*t

When they pass, the position is the same (obviously), so

(100/60)*t=300-(150/60)*t

Re-arrange that, and you get:-

( (100/60) + (150/60) ) * t = 300

Divide through by ( (100/60) + (150/60) ) and you get

t = 300 / ( (100/60) + (150/60) )

Which comes out at 72 minutes by my calculator.

So, they'll pass each other at 11:12.

And if you want the position, just use aircraft A: (100/60)*72 = 120nm. So they meet 120Nm from airport A.

To double check, put aircraft B's equation in: 300 - (150/60)*72 = 120, same answer.

G
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