Question 418102
Let {{{t}}} = time for CHI to NY trip
(total distance) / (total time) = (average speed) 
given:
Time for NY to CHI trip is {{{900 / 30 = 30}}} hrs
{{{ (2*900) / (30 + t) = 60}}}
{{{ 1800 = 60*(30 + t) }}}
{{{1800 = 1800 + 60t}}}
{{{60t = 0}}}
{{{t = 0}}}
In order to average 60 mi/hr for the 1800 mi trip,
the driver must make it back in 0 hours. That 
means he can't do it. So maybe 60 mi/hr is the upper
limit.  What if he wants to average 59 mi/hr?
Can he do it?
{{{ 1800 / (30 + t) = 59}}}
{{{ 1800 = 59*(30 + t)}}}
{{{1800 = 1770  + 59t}}}
{{{59t = 30}}}
{{{t = 30/59}}}
{{{t = .508}}} 
Yes, but he'll have to make it back in about 1/2 hr
His average speed on the return trip would be:
{{{900/.508 = 1771.65}}} mi/hr
Well, no cops could catch him if he survived the 
acceleration and deceleration.
He would have to average quite a bit lower
speed on the return trip.