SOLUTION: Jim and John drive from point A to point B in 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 spe

Jim covered the distance from A to B in 10 hours, according to the condition.

John covered the same distance in 5 hours,  according to the condition.


Hence, John' speed is two times Jim' speed.


OK.  Let  u  be the Jim' speed.

     Then John' speed is (2u).


Let's write the equation/equations of theirs movements.


Let "t" be the time counted from Jim' start at 6:00 am.


Then Jim's distance from A at the time moment t is  x = u*t, 

while John's distance from A at the time t is y = (2u)*(t-4).


At the catching moment  x = y,  or


ut = (2u)*(t-4),

ut = 2ut - 8u  ====>  ut = 8u  ====>  t = 8.


So, John will catch up Jim  8 hours after  6:00 am, i.e. at 2:00 pm.


Answer.  John will catch up Jim  at 2:00 pm.