A man travels to Austin, Texas at 40 mph and returns, on the same route, at 60 mph. What is his average speed? Thank you.
+33653 Let the distance between where he starts from and Austin be L miles
Time taken to get to Austin = L/40 hours
Time taken to return from Austin = L/60 hours
Average speed = total distance/total time = 2L/(L/40 + L/60) → 2/(1/40 + 1/60) → 48 mph
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+33653 You are to be commended for trying owlface.
You are not the only person to make this mistake when averaging speeds. Even the Guiness Book of Records gets it wrong when calculating land speed record attempts!! They take the speeds, calculated seperately, over two distances (there and back) and average them in just the way you did to get an overall average speed. What they get by doing this is not the true average (unless the speeds happen to be the same in both directions)!
+37084 Agree with discussion.....ONLY when TIME is the same will the average of the two speeds be the simple average of the speeds added together.
40 mph for 1 hr
60 mph for 1 hr average speed is 50 (because you travelled (40 miles + 60 miles)/ (2 hours) )
+12530 A man travels to Austin, Texas at 40 mph and returns, on the same route, at 60 mph. What is his average speed?
Let everybody chime in on this one and will see how many DIFFERENT solution can be had!!.
Here is my take:
The man travels 1 mile in 1/40 of an hour going, and in 1/60 of an hour coming:
Therefore: 1/2 [1/40 + 1/60] =1/48 of an hour, going and coming.
Therefore he will average the reciprocal of that, or 48 hours.
Let everybody chime in on this one and will see how many DIFFERENT solution can be had!!.
Here is my take:
The man travels 1 mile in 1/40 of an hour going, and in 1/60 of an hour coming:
Therefore: 1/2 [1/40 + 1/60] =1/48 of an hour, going and coming.
Therefore he will average the reciprocal of that, or 48 hours.
