posted by howard on .
can you help i dnt know what to do so could you tell me how to solve it
1. every weekend, Phil drives from his home to his favorite amusement park, a distance of 60 miles.
a. If he averages 40 mph for the first half of the trip, what must his average speed for the second half of the trip in order for him to average 50 mph for the entire trip?
b. Suppose that on the trip home, Phil decides to stop a 30 minute dinner break. If he averages 45 mph for the first half of this return trip and 60 mph for the second half of this trip, how many more MINUTES will the return trip take compared to the trip going to the park? (be sure to account for his dinner break.)
1.a To average Vav = 50 mph for the entire trip, the 60 miles distance (D) must be covered in a total time of
D/Vav = 6/5 = 1.2 hours
The first 30 miles required 30/40 = 0.75 hours. The remaining 30 miles must be completed in 1.2 -0.75 = 0.45 hours.
The average speed for the second half of the trip must be 30/0.45 = 66.7 mph
1.b The trip home is also 60 miles in length. Total travel time with dinner break = 30/45 + 30/60 hrs + 0.5 h = 1.667 h
Remember that it took him 1.2 hours to get TO the park. With the dinner break and return speeds indicated, it takes 0.467 hours longer to get home.