A Awesome31312 Nov 2014 71 0 Home Jul 30, 2016 #1 It goes like this: "Ed goes 20mph in one direction, and 50 mph on the return trip. His average speed is?" I am just not getting the explanation in the book, if anybody could explain how this is done, I would be extremely grateful, thanks

It goes like this: "Ed goes 20mph in one direction, and 50 mph on the return trip. His average speed is?" I am just not getting the explanation in the book, if anybody could explain how this is done, I would be extremely grateful, thanks

skeeter MHF Helper Jun 2008 16,217 6,765 North Texas Jul 30, 2016 #2 average speed = (total distance traveled)/(total time required to cover the total distance) let d = distance in one direction $\implies$ total distance = 2d time traveling at 20 mph = d/20 time traveling at 50 mph = d/50 total time = d/20 + d/50 = 7d/100 avg speed = (2d)/(7d/100) = (200d)/(7d) = 200/7 mph $\approx$ 28.6 mph Reactions: 1 person

average speed = (total distance traveled)/(total time required to cover the total distance) let d = distance in one direction $\implies$ total distance = 2d time traveling at 20 mph = d/20 time traveling at 50 mph = d/50 total time = d/20 + d/50 = 7d/100 avg speed = (2d)/(7d/100) = (200d)/(7d) = 200/7 mph $\approx$ 28.6 mph

D DenisB Feb 2015 2,255 510 Ottawa Ontario Jul 30, 2016 #3 ...or make up li'l scenario: @20............100.............>5 2<[email protected] 200 miles travelled in 7 hours, right?

...or make up li'l scenario: @20............100.............>5 2<[email protected] 200 miles travelled in 7 hours, right?