How to solve time/distance equation problems

[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?