A bus leaves a station at 1 p.m., traveling west at an average 
rate of 44 mi/h.  One hour later a second bus leaves along the 
same route, traveling at 54 mi/h.  At what time will the two 
buses be 274 mi apart?

Wow!!!!!!  That'll sure take a long time with the second bus 
traveling only 10 mi/hr faster than the first bus!!!!!!! At
only 10 mi/hr faster, the second bus won't even catch up to
the first to pass it until 4.4 hours later at 6:24 PM.  Then
to get 274 miles ahead will take more than a whole day.  Are
you sure you have this one copied right?  It's a ridiculous
problem, but it can be imagined. So I'll do it anyway.  Let's
assume they started on Monday.  It will be at least Tuesday 
evening before the 2nd bus is 274 miles ahead of the first bus,
going only 10 mph faster.  

Make this chart

             D      R    T   
1st bus|    
2nd bus| 

Let t be the time the first bus traveled.
Then since the 2nd bus started an hour later
its traveling time is t-1.  So fill in the 
times

             D      R    T   
1st bus|                 t
2nd bus|                t-1

Fill in the given rates (speeds):

             D      R    T   
1st bus|           44    t
2nd bus|           54   t-1

Use D = RT to fill in the distances:

             D      R    T   
1st bus|    44t    44    t
2nd bus|  54(t-1)  54   t-1

Now the distance the 2nd bus traveled
is 274 miles more than the 1st bus

So we form the equation by

2nd bus's distance = 1st bus's distance + 274

   54(t-1) = 44t + 274 

Answer = 32.8 hours

So if the first bus started at 1 p.m on Monday
the second bus will be 274 miles ahead of the
1st bus at 9:48 p.m Tuesday night.

I think you copied the problem wrong.

Edwin