Travel Word Problem

Question

Jay traveled from Miami to Jacksonville, a distance of 320 miles. He left Miami at 8:00 A.M. stopped for lunch for 30 minutes, and stopped for two 15 minute breaks at rest areas. He arrived at Jacksonville at 2:15 p.m. What was the average rate of speed during his travel time

Answer 1

distance=rate*time

he lost 1 hour in traveling time (30+15+15)

8:00 AM to 2:15 PM is 6 hours and 15 minutes minus 1 hour equals 5 hours 15 minutes

5 hours 15 minutes is 5 1/4 hours

d-r*t

320=r*(5.25)

r=320/5.25

r=60.95238 miles per hour

r≈61 mph