Distance/speed/time?
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Distance/speed/time?
Hi, ive got my ctc coming up, just revising my long division etc etc and just came across a question that i cant get round my head to answer! its as follows:
aircraft a sets off from airport a at 10am and flies towards airport b with a flying speed of 100kts.
aircraft b sets off from airport b at 10am and flies towards airport a with a flying speed of 150kts.
airports a and b are 300miles apart
at what time do the aircraft meet each other?
Is this a simple distance/speed/time answer, im sure its easy, but havent done this kind of maths for years! If anyone could give me a helping hand id appreciate it!
thanks!
aircraft a sets off from airport a at 10am and flies towards airport b with a flying speed of 100kts.
aircraft b sets off from airport b at 10am and flies towards airport a with a flying speed of 150kts.
airports a and b are 300miles apart
at what time do the aircraft meet each other?
Is this a simple distance/speed/time answer, im sure its easy, but havent done this kind of maths for years! If anyone could give me a helping hand id appreciate it!
thanks!
Last edited by TicTac1982; 24th Jul 2004 at 15:19.
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
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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They have a combined speed of 250kts.
The distance they are travelling is 300nm
time=300/250
simplifies to 6/5 which is 1 1/5
so 1hr and 12mins
same answer
am i right in going along those lines as you are expected to do the calculation in your head without a calculator.
The distance they are travelling is 300nm
time=300/250
simplifies to 6/5 which is 1 1/5
so 1hr and 12mins
same answer
am i right in going along those lines as you are expected to do the calculation in your head without a calculator.
