Question 157645
OK
Distance(d) equals Rate(r) times Time(t) or d=rt;  r=d/t and t=d/r

So, before Dave gets started, Alan has already driven 55*1 or 55 miles (and they have talked an hour on their phones)

Let t=time it takes them to get 250 mi apart after Dave leaves

Total distance Alan drives 55+55t
Total distance Dave drives=65t

Now when the above two distances add up to 250 mi, they will be at the outer limits of phone reception, so:
55+55t+65t=250 subtract 55 from each side
55-55+65t+55t=250-55  collect like terms
120t=195  divide each side by 120
t= 1 5/8 hr or 1hr 37.5 min--------time it takes to get 250 mi apart after Dave leaves

Time that they can talk on the phone is the 1 hour that Alan was driving alone plus the 1 5/8 hour that the were both driving and this equals 2 5/8 hrs


Hope this helps---ptaylor