You can do this in your head, but I'll do it by
algebra as well. 

How to do it in your head:

at 11AM, when John leaves, Martina is already 
48 miles down the road, since she has pedaled 
at the (break-neck) speed of 24 mi/hr on a 
bicycle for two hours!  So John, driving at 
48 mi/hr is approaching her at 24 mi/hr, so he 
has to make up her 48 mile lead at 24 mi/hr
approach rate, so it'll obviously take him 2 
hours, and 2 hours past 11AM is 1PM.

How to do it by algebra.  Make this chart:

            DISTANCE     RATE     TIME
Martina                                 
John                                   

Let the answer be t. So write t for John's 
catch-up time


            DISTANCE     RATE     TIME
Martina                               
John                                t

Since Martina has a 2-hour head start, her 
time is t+2. So fill that in:


            DISTANCE     RATE     TIME
Martina                            t+2
John                                t

Now fill in their rates which are given as 
24 mi/hr and 48 mi/hr

            DISTANCE     RATE     TIME
Martina                   24       t+2
John                      48        t


Now use DISTANCE = RATE×TIME to fill in the 
two DISTANCEs:

            DISTANCE     RATE     TIME
Martina      24(t+2)      24       t+2
John           48t        48        t

Now since they both traveled the same distance,
we set the DISTANCEs equal:

       24(t+2) = 48t
     
Solve that and get t = 2 hours, which means John, 
starting two hours later than 9AM, or 11AM, will 
catch up to her in 2 hours past 11AM, or 1PM.

Edwin