Question 339641
going 15 mph he arrives 2 hrs sooner than if he was going 10 mph


15(t) = 10(t + 2) ___ 15t = 10t + 20 ___ 5t = 20 ___ t = 4


so the distance is 15(4) or 60 mi


he wants to arrive an hour later than a 15 mph trip , so ___ t = 4 + 1 = 5


a 5 hr trip , arriving at 5:00 PM ___ he should leave at noon


60 mi in 5 hr is 12 mph  (60/5)