Question 360345
Jim and John drive from point A to point B  separate cars. Jim leaves at 6 am and arrives at 4 pm. John leaves at 10 am and arrives at 3 pm. Assume both men drive at constant speeds. Find when John catches up with Jim.
--------------
m = Jim's speed
n = John's speed
---------
d = rt
To go the total distance:
d = m*10 --> m = d/10
d = n*5 --> n = d/5
-------
Jim starts at 0600, t = 0
At some time t:
m*t = n*(t-4)
(d/10)t = (d/5)*(t-4)
d cancels
t/10 = (t-4)/5
t = 2(t-4)
t = 2t - 8
t = 8 hours
0600 + 8 = 1400 or 2 PM
-----------------
Interesting problem.
-------------------
Another approach:
John's speed is 2x Jim's speed.  Since John arrived 1 hour before Jim going 2x as fast, John covered the same distance in his last hour that Jim covered in 2 hours.
--> They were at the same point 1 hour before John's arrival and 2 hours before Jim's arrival.
--> 3PM - 1 hour = 4PM - 2 hours
= 2PM