Aptitude Test Question

[TDK90]

Can anybody help me with this type of question:

AC 1 departs ABC at 1400Z TAS 140 kts
AC 2 departs ABC at 1600Z TAS 320 kts

How long till AC 2 overtakes AC 1?

Is there a generic formula to use for these questions?

 

[five-alive]

Distance=rate*time

D1 = 140 * t
D2 = 320 * t - 640

The (-) 640 resets the distance to 0 when t=2 for D2.

The overtake will occur when the distances are equal

D1 = D2

t=3.55555

Just algebra after that. No generic formula, just some reasoning and math


ANS

 

[airspeedsalive]

Damn. I guess I am as dumb as my teachers thought.

 

[AK47]

Distance=rate*time

D1 = 140 * t
D2 = 320 * t - 640

The (-) 640 resets the distance to 0 when t=2 for D2.

The overtake will occur when the distances are equal

D1 = D2

t=3.55555

Just algebra after that. No generic formula, just some reasoning and math


ANS

The correct answer is

140 x (t + 2) = 320 * t

or

180 * t = 280

or

t = 280 / 180 = 1.555 hours or 1 hour 33 minutes 20 seconds

 

[charter dog]

Or looked at another way: (for those that prefer simple figurin'!)

Aircraft 1 has a 280 mile head start. (140 X 2) Aircraft two is 180 knots faster. (320 - 140) Time to make up 280 nm at 180 kts. of overtake = 280/180 = 1.555 hours or 1.555 X 60 = 93 minutes and 20 seconds after aircraft 2 begins. Overtake will occur at a time 3.555 hours after aircraft 1 begins and 1.555 hours after aircraft 2 begins. 3.555 X 140 = 497.7 nm and 1.555 X 320 = 497.6 nm. Close enough!

Best,

 

[flatspin7]

The real solution is that if you are flying Aircraft B just climb up to the next altitude so you really dont have to care when you will over take him!

 

[SEVEN]

I hate math!

 

[TDK90]

Or looked at another way: (for those that prefer simple figurin'!)

Aircraft 1 has a 280 mile head start. (140 X 2) Aircraft two is 180 knots faster. (320 - 140) Time to make up 280 nm at 180 kts. of overtake = 280/180 = 1.555 hours or 1.555 X 60 = 93 minutes and 20 seconds after aircraft 2 begins. Overtake will occur at a time 3.555 hours after aircraft 1 begins and 1.555 hours after aircraft 2 begins. 3.555 X 140 = 497.7 nm and 1.555 X 320 = 497.6 nm. Close enough!

Best,

Thanks! That's exactly the kind if math I like...

 

[five-alive]

The correct answer is

140 x (t + 2) = 320 * t

or

180 * t = 280

or

t = 280 / 180 = 1.555 hours or 1 hour 33 minutes 20 seconds

We basically came up with the same answers just a different time benchmark, 3.555 hrs from the time the first plane left, or 1.555 from the time the second plane left

 

[AC560]

I can only do these types of problems if they involve 2 trains and 1 or more train stations.

 

[User998]

The real solution is that if you are flying Aircraft B just climb up to the next altitude so you really dont have to care when you will over take him!

Don't forget to figure in the 10 degree to the right vectors for overtaking traffic. That may change things up a minute or two.

 

[Gutenberg]

sounds like a GMAT question. draw the rate pie!!!

 

[kingsize]

I did a search but couldn't find the ICAO code ABC. Can someone find that out? Then I'll be able to work the math problem. Thanks.