How to solve time/distance equation problems

[Ratatosk]

I guess you have all seen this type of question before? How do I solve it?

An aircraft with a speed of 250kts is overhead point E at 06:30. Another aircraft with a speed of 300kts follows the first aircraft and passes E at 06:36. At what distance from E will the faster aircraft catch up with the slower?

I'm having a hard time with math so please be as specific as possible (don't leave out "the obvious" steps please) and any shortcuts avoiding the equation all together would be greatly appreciated :) How to think, what is what and why when solving it?

Thanks!

[Tinstaafl]

Aircraft A 250 kts
Aircraft B 300 kts
----------------------------------
Speed difference: B is 50 kts faster than A.


A crosses the fix at 0630
B crosses the fix at 0636
---------------------------------------
B is 6 minutes behind A.

At the time B crosses the fix A will have travelled 6 mins at 250 kts = 25 nm

When will B catch A? 25 nm to catch at 50 kts speed difference = 30 mins

Where will B catch A?
30 mins after A's 0636 position at 250 kts = 125 nm further on = 125 nm + 25 nm = 150 nm from the fix.

Check correctness:
30 mins after B's 0636 position at 300 kts = 150 nm from the fix which agrees with A's position above.

[Curtis E Carr]

At 0636, when the faster aircraft passes Point E, the slower aircraft is already 25 nm beyond Point E. The difference in speed between the two aircraft means that the faster aircraft is closing with the slower one at a rate of 50 kts, or 5nm per 6 minutes. This means that the gap will be closed in (25/5*6) or 30 minutes. In 30 minutes after Point E (i.e. at 0706hrs), the faster aircraft will have travelled 150 nm.

You can then double check by calculating where the slower aircraft would also be at 0706 hrs by a simple time and distance calculation (i.e. 250 kts travelled in 30 minutes equals 125 nm plus the 25 miles already travelled beyond Point E between 0630 and 0636 brings the answer to 150 nm).

[Keith.Williams.]

The two aircraft will pass when they have both flown the same distance from the starting point.


Let’s use T to represent the time that this occurs. measured from the time at which the first aircraft passes the initial positiion (06:30)


Distance flown = time x Speed


So at Time T aircraft A has flown for T hours at 250 nm/hour.


So it has flown 250T nm.



Aircaft B leaves the start point at 06:36.


This is 6 minutes later than aircraft A.


Dividing 6 minutes by 60 minutes/hour gives a time of 0.1 hours.


So aircraft B starts out 0.1 hour after aicraft A.


So when time = T, aircraft B has been flying for (T – 0.1 hours)


So distance flown by aircraft B = (T – 0.1) x 300 nm/hour.


Which is 300 T – 30 nm.


But the two aircraft are now in the same position, so they have both flown the same distance.


So 250 T = 300 T – 30.


So 30 = 300T – 250 T


So 30 = 50 T

So 30/50 = T


So T = 0.6 hours.


Multiplying this by 60 minutes/hour gives 356 minuites.


So the two airacrft pass 36 minute safter aircarft A starts out.


That is at 06:30 + 36 = 07:12.


To test the answer:

At time T aircraft A has flown 0.6 hours x 250 kts = 150 nm
At time T aircraft B has flown (0.6 – 0.1) hours x 2 300 kts = 150 nm.

This is a rather long explanation, but you did ask for all of the details?

[Ratatosk]

Brilliant! Thanks a bunch guys! I knew there was a way to simply reason out the solution!